Download - TRIGONOMETRY One mark questions
TRIGONOMETRY
One mark questions
1. Define an angle.
Solution: An angle is an amount of rotation of a half-line in a
plane about its end from its initial position to a terminal
position.
2. What do you mean by measure of an angle?
Solution: The amount of rotation from initial position to
terminal position is called measure of an angle.
3. Define one radian.
Solution: A Radian is the angle subtended at the centre of a
circle by an arc, whose length is equal to its radius. 1 radian is
denoted by 1π. One radian is also called one circular measure.
4. Convert the following into radian measure:
a) 18Β° b) 7.5Β° c) 105Β° d) 1
2
Β° e) 36Β°32β²34β²β²
Solution:
a) 1Β° = Ο
180
c β΄ 18Β° =
Ο
180Γ 18
c=
Ο
10
c
b) 7.5Β° = Ο
180Γ 7.5
c=
Ο
180Γ
15
2
c=
Ο
24
c
c) 105Β° = Ο
180Γ 105
c=
7Ο
12
c
d) 1
2
Β°=
Ο
180Γ
1
2
c=
Ο
360
c
e) 36Β°32β²34β²β² = 36 +32
60+
34
3600 Β°
= 36.5427 Β° = 36.5427 Γ 0.0174 c
(since 1Β° = 0.0174 c) β΄ 36Β°32β²34β²β² β 0.6358 radian.
5. Express the following in degree measure:
a) π
36 b)
7π
6 c)
23π
4 d)
7π
24 e)
5π
12
Solution:
a) 1π = 180
π Β° β΄
π
36=
180
πΓ
π
36 Β°
= 5Β°
b) 7π
6=
180
πΓ
7π
6 Β°
= 210Β°
c) 23π
4=
180
πΓ
23π
4 Β°
= 1035Β°
d) 7π
24=
180
πΓ
7π
24 Β°
= 105
2 β
= 52.5 Β°
e) 5π
12=
180
πΓ
5π
12 Β°
= 75β.
6. Prove that π ππ π΄ β πππ‘ π΄ = πππ π΄
Solution: πΏπ»π = π ππ π΄ β πππ π΄
π ππ π΄= πππ π΄ = π π»π.
7. Prove that πππ‘ π΄ β π ππ π΄ β sin π΄ = 1
Solution: πΏπ»π = πππ π΄
π ππ π΄β
1
cos π΄β π ππ π΄ = 1 = π π»π.
8. Prove that πππ‘2 π β 1 β πππ 2 π = πππ 2 π
Solution: πΏπ»π = πππ 2 π
π ππ 2 πβ π ππ2 π = πππ 2 π.
9. Prove that πππ ππ πΌ β 1 β π ππ2 πΌ = cot πΌ
Solution: πΏπ»π = 1
π ππ πΌβ πππ πΌ = πππ‘ πΌ = π π»π.
10. Prove that (1 + π‘ππ2 π). 1 β πππ 2 π = π‘ππ2 π
Solution: πΏπ»π = π ππ2 π β π ππ2 π = 1
πππ 2 πβ π ππ2 π = π‘ππ2 π = π π»π.
11. Prove that (1 + πππ‘2 π΄) β π ππ2 π΄ = 1
Solution: πΏπ»π = πππ ππ2 π΄ β π ππ2 π΄
= πππ ππ π΄ β π ππ π΄ 2 = 12 = 1 = π π»π.
12. Prove that πππ πΌ β πππ ππ πΌ β π ππ2 πΌ β 1 = 1.
Solution: πΏπ»π = πππ πΌ β1
π ππ πΌβ π‘ππ2 πΌ
= πππ πΌ β1
π ππ πΌβ π‘ππ πΌ =
πππ πΌ
π ππ πΌβπ ππ πΌ
πππ πΌ = 1 = π π»π.
13. Prove that πππ π β πππ‘2 π β 1 = πππ ππ2 π β 1.
Solution: πΏπ»π = cos π β πππ ππ π
= cos π β1
π ππ π = πππ‘ π = πππ ππ2 π β 1 = π π»π.
14. Prove that π ππ2 π΄ β πππ‘2 π΄ + π ππ2 π΄ = 1.
Solution: πΏπ»π = π ππ2 π΄ βπππ 2 π΄
π ππ 2 π΄ + π ππ2 π΄
= πππ 2 π΄ + π ππ2 π΄ = 1 = π π»π.
15. Define coterminal angles. Give two examples.
Solution: If the initial and terminal sides of two angles are
equal then the angles are called coterminal angles.
16. Prove that πππ ππ (180Β° β π) = πππ ππ π.
Solution: πππ ππ (180Β° β π) = 1
π ππ (180Β° β π) =
1
π ππ π = πππ ππ π.
17. Prove that π ππ (180Β° β π) = β π ππ π.
Solution: π ππ (180Β° β π) = 1
πππ (180Β° β π) =
1
β πππ π = β π ππ π.
18. Prove that πππ‘ (180Β° β π) = β πππ‘ π.
Solution:πππ‘ (180Β° β π) = 1
π‘ππ (180Β° β π) =
1
β π‘ππ π = β πππ‘ π.
19. Prove that πππ ππ (180Β° + π) = β πππ ππ π.
Solution: πππ ππ (180Β° + π) = 1
π ππ (180Β° + π) =
1
β π ππ π = β πππ ππ π.
20. Prove that π ππ (180Β° + π) = β π ππ π.
Solution: π ππ (180Β° + π) = 1
πππ (180Β° + π) =
1
β πππ π = β π ππ π.
21. Prove that πππ‘ (180Β° + π) = β πππ‘ π.
Solution: πππ‘ (180Β° + π) = 1
π‘ππ (180Β° + π) =
1
β π‘ππ π = β πππ‘ π.
22. Prove that πππ ππ (270Β° β π) = β π ππ π.
Solution: πππ ππ (270Β° β π) = 1
π ππ (270Β° β π) =
1
β πππ π = β π ππ π.
23. Prove that π ππ (270Β° β π) = β πππ ππ π.
Solution: π ππ (270Β° β π) = 1
πππ (270Β° β π) =
1
β π ππ π = β πππ ππ π.
24. Prove that πππ‘ (270Β° β π) = π‘ππ π.
Solution: πππ‘ (270Β° β π) = πππ‘ (180Β° + 90Β° β π)
= πππ‘ (90Β° β π) = π‘ππ π.
25. Prove that πππ ππ (270Β° + π) = β π ππ π.
Solution: πππ ππ (270Β° + π) = 1
π ππ (270Β° + π) =
1
β πππ π = β π ππ π.
26. Prove that π ππ (270Β° + π) = πππ ππ π.
Solution: π ππ (270Β° + π) = 1
πππ (270Β° + π) =
1
π ππ π = πππ ππ π.
27. Prove that πππ‘ (270Β° + π) = β π‘ππ π.
Solution: πππ‘ (270Β° + π) = πππ‘ (180Β° + 90Β° + π)
= πππ‘ (90Β° + π) = β π‘ππ π.
28. Define periodic function.
Solution: A function π(π₯) is said to be periodic if
π( π₯ + π) = π(π₯)
for some real number π.
If π is the smallest positive real number satisfying
π( π₯ + π) = π(π₯) then π is called the period of π(π₯).
29. Write the period of all the trigonometric functions (1 mark
each).
Solution:
Period of π ππ π₯ is 2π ; Period of πππ π₯ is 2π ;
Period of π‘ππ π₯ is π ; Period of πππ‘ π₯ is π ;
Period of πππ ππ π₯ is π ; Period of π ππ π₯ is π.
30. Prove that π ππ 45Β° + π΄ =1
2(π ππ π΄ + πππ π΄)
Solution: π ππ 45Β° + π΄ = π ππ 45Β°. πππ π΄ + πππ 45Β°. π ππ π΄
= 1
2πππ π΄ +
1
2π ππ π΄) =
1
2(π ππ π΄ + πππ π΄).
31. Find the value of π ππ 54Β°
Solution: π ππ 54Β° = π ππ (90Β° β 36Β°) = πππ 36Β° = 5 + 1
4.
32. Find the value of πππ 72Β°
Solution: πππ 72Β° = πππ (90Β° β 18Β°) = π ππ 18Β° = 5 β 1
4.
33. Prove that 2 π ππ 15Β° πππ 15Β° =1
2.
Solution: 2 π ππ 15Β° πππ 15Β° = π ππ 2 β 15Β° = π ππ30Β° =1
2
34. Prove that π ππ π
8πππ
π
8=
1
2 2
Solution: πΏπ»π = 2
2. π ππ
π
8. πππ
π
8 =
1
2 2π ππ
π
8. πππ
π
8
= 1
2π ππ 2 β
π
8 =
1
2π ππ
π
4 =
1
2Γ
1
2 =
1
2 2.
35. Prove that π ππ π₯
1 + πππ π₯= π‘ππ
π₯
2
Solution: π ππ π₯
1 + πππ π₯ =
2π ππ π₯
2.πππ
π₯
2
2πππ 2 π₯
2
= π ππ
π₯
2
πππ π₯
2
= π‘ππ π₯
2.
36. Prove that 1 + πππ 2π₯
1 β πππ 2π₯ = πππ‘ π₯
Solution: 1 + πππ 2π₯
1 β πππ 2π₯ =
2πππ 2 π₯
2π ππ 2 π₯ =
πππ π₯
π ππ π₯ = πππ‘ π₯.
37. Express 2 π ππ2π΄ β πππ π΄ as a sum.
Solution: 2 π ππ 2π΄. πππ π΄ = π ππ 2π΄ + π΄ + π ππ 2π΄ β π΄
= π ππ 3π΄ + π ππ π΄.
38. Express πππ 7π β πππ 5π as a sum.
Solution: πππ 7π β πππ 5π = 1
2 πππ 7π + 5π + πππ 7π β 5π
= 1
2 πππ 12π + πππ 2π .
39. Express π ππ 4π΄ + π ππ 2π΄ as product.
Solution: π ππ 4π΄ + π ππ 2π΄ = 2 π ππ 4π΄ + 2π΄
2 . πππ
4π΄ β 2π΄
2
= 2 π ππ 3π΄ . πππ π΄.
40. Express πππ 60Β° β πππ 20Β° as a product.
Solution: πππ 60Β° β πππ 20Β° = β 2 π ππ 60 Β° + 20 Β°
2 . π ππ
60Β° β 20Β°
2
= β 2 π ππ 40Β° β π ππ 20Β°.
41. Express πππ 55Β° + π ππ 55Β° as a product.
Solution: πππ 55Β° + π ππ 55Β° = πππ 55Β° + πππ 90Β° β 55Β°
= πππ 55Β° + πππ 35Β° = 2 πππ 55 Β° β 35 Β°
2 . πππ
55Β° β 35Β°
2
= 2 πππ 45Β° . πππ 10Β° = 2.1
2. πππ 10Β° = 2. πππ 10Β°.
42. Define a trigonometric equation.
Solution: An equation involving one or more trigonometric
functions are called trigonometric equation.
43. What do you mean by general solution of a trigonometric
equation.
Solution: The set of all values of the angle, π (involved in a
trigonometric equation) satisfying the trigonometric equation,
π(π) = 0 is called general solution of the trigonometric
equation.
44. Find the general solution of π ππ π = 0.
Solution: When = 0 , Β± π , Β± 2π , Β± 3π, β¦ π ππ π = 0.
β΄ the general solution of π ππ π = 0 is π = ππ, π β π.
45. Find the general solution of πππ π = 0.
Solution: When π = Β± π
2, Β±
3π
2, Β±
5π
2 ,β¦ πππ π = 0.
β΄ the general solution of πππ π = 0 is π = (2π + 1)π
2 , π β π.
46. Find the general solution of π‘ππ π = 0.
Solution: When π = 0 , Β± π , Β± 2π , Β± 3π, β¦ π‘ππ π = 0.
β΄ the general solution of π‘ππ π = 0 is π = ππ, π β π.
47. Find the principal value of πππ π₯ = 3
2
Solution: πππ π₯ = 3
2> 0 β΄ π₯ lies in the I or IV quadrant.
Principal value of π₯ β 0, π . Since πππ π₯ is positive, the principal
value is in the I quadrant.
πππ π₯ = 3
2= πππ
π
6 and
π
6β 0, π . β΄ principal value of π₯ is
π
6.
48. Find the principal value of π ππ π₯ =1
2.
Solution: π ππ π₯ =1
2> 0 β΄ π₯ lies in the I or II quadrant.
Principal value of π₯ β βπ
2,
π
2 . Since π πππ₯ is positive the
principal value is in the I quadrant.
π πππ₯ =1
2= π ππ
π
6 and
π
6β β
π
2,
π
2 . β΄ principal value of π₯ is
π
6.
49. Find the principal value of πππ ππ π = β2
3.
Solution: πππ ππ π = β 2
3 βΉ π ππ π = β
3
2< 0
β΄ π₯ lies in the III or IV quadrant.
Principal value of π β βπ
2,
π
2 . Since π πππ is negative the
principal value is in the IV quadrant.
π ππ π = β 3
2= π ππ β
π
3 and β
π
3 β β
π
2,
π
2 .
β΄ principal value of π₯ is βπ
3.
50. Find the principal value of πππ π₯ = β 3
2.
Solution: πππ π₯ = β 3
2< 0 β΄ π₯ lies in the II or III quadrant.
Principal value of π₯ β 0, π . Since πππ π₯ is negative the
principal value is in the II quadrant.
πππ π₯ = β 3
2= β πππ
π
6= πππ 180Β° β 30Β°
= πππ 150Β° = πππ 5π
6 and
5π
6β 0, π .
β΄ principal value of π₯ ππ 5π
6.
Two mark questions
1. Define sexagesimal system.
Solution: In this system, a right angle is divided into 90 equal
parts. Each part is equal to one degree. One degree is divided
into 60 equal parts. Each part is called one minute. One minute
is divided into 60 equal parts. Each part is called one second.
One right angle = 90 degrees, written as 90Β° ;
one degree, 1Β° = 60 minutes, written as 60β² ;
one minute, 1β² = 60 seconds, written as 60β²β²;
2. Prove that one radian is a constant angle.
Solution: Since the angle subtended at the center of a circle of
unit radius, by an arc of length one unit is one radian, the
angle subtended at the center by the circumference of the
circle is 2π radian. Thus 2ππ = 360Β°.
β΄ ππ = 180Β°.
β΄ 1π =180
π.
This gives the conclusion, 1π is a constant angle.
3. With usual notations prove that π = ππ.
Solution: We know that in a circle of radius π, an arc of
length π, subtends an angle of one radian at the center of the
circle. Since the angle subtended at the center by an arc of
length π is one radian, the angle subtended by an arc of length
π has the measure = π
π. Thus if π is the angle subtended at the
center by an arc of length π then π
π = π β΄ π = ππ.
4. With usual notations prove that the area of a sector of a
circle is given by 1
2π2π or
1
2ππ
Solution: We know that in a circle of radius π, the angle 2π
radian traces the area ππ2. Therefore the area of the sector
tracing an angle π radian at the center is π΄ =ππ2
2πΓ π =
1
2π2π.
Since π = ππ we get π΄ =1
2π2π =
1
2π. ππ =
1
2ππ.
5. An arc of a circle subtends 15Β° at the centre. If the radius is
4 cm, find the length of the arc and area of the sector formed.
Solution: Here π = 4 ππ and π = 15Β° = 15 Γπ
180=
π
12
β΄ π = ππ βΉ π = 4 Γ22
7Γ
1
12=
22
21 ππ.
Area of the sector =1
2ππ =
1
2 Γ 4 Γ
22
21 =
44
21 π π. ππ.
6. An arc of length 11
3 cm subtends an angle of 30Β° at the
centre of a circle. Find the area of the sector of the circle so
formed.
Solution: Given =11
3 , π = 30Β° =
π
6.
We have π = ππ βΉ π =π
π=
11
3Γ
6
π=
11
3Γ
6
22Γ 7 = 7ππ
π΄ =1
2 π2π =
1
2 7 2 π
6=
1
2 Γ 49 Γ
22
7Γ
1
6=
77
6 π π. ππ.
7. If, in two circles arcs of the same length subtend angles of
60Β° πππ 75Β° at the centre, find the ratio of their radii.
Solution: Let π1 and π2 be the radii of the two circles.
Given π1 = 60Β° = π
180Γ 60 =
π
3 and π2 = 75Β° =
π
180Γ 75 =
5π
12
Now the length of each of the arc, π = π1π1 = π2π2, which gives
π1 Γπ
3= π2 Γ
5π
12βΉ
π1
π2=
5
4. Hence π1: π2 = 5: 4.
8. Prove the following:
a) π ππ π. πππ ππ π = 1, πππ ππ π = 1
π ππ π ;
b) πππ π. π ππ π = 1, π ππ π = 1
πππ π ;
c) π‘ππ π. πππ‘ π = 1, πππ‘ π = 1
π‘ππ π ;
d) π ππ π
πππ π = π‘ππ π; e)
πππ π
π ππ π = πππ‘ π.
f) π ππ2π + πππ 2π = 1;
g) 1 + π‘ππ2π = π ππ2π ; h) 1 + πππ‘2π = πππ ππ2π.
Solution: Consider a circle of
radius π centered at the origin of
the Cartesian coordinate system. Let
π(π₯, π¦) be a point on the circle. Join
ππ. Define angle π having ππ has
the initial direction and ππ as the
terminal direction. Consider a circle
of radius π centered at the origin of
the Cartesian coordinate system. Let π(π₯, π¦) be a point on the
circle. Join ππ. Define angle π having ππ has the initial
direction and ππ as the terminal direction. Then
π ππ π = π¦
π , ππ π =
π₯
π , π‘ππ π =
π¦
π₯ ,
πππ‘ π = π₯
π¦, π ππ π =
π
π₯, πππ ππ π =
π
π¦.
a) π ππ π. πππ ππ π = π¦
π.π
π¦ = 1. β΄ πππ ππ π =
1
π ππ π ;
b) πππ π. π ππ π = π₯
π. π
π₯ = 1. β΄ π ππ π =
1
πππ π ;
c) π‘ππ π. πππ‘ π = π¦
π₯.π₯
π¦ = 1. β΄ πππ‘ π =
1
π‘ππ π ;
d) π ππ π
πππ π =
π¦ π
π₯ π =
π¦
π₯ = π‘ππ π ;
e) πππ π
π ππ π =
π₯ π
π¦ π =
π₯
π¦ = πππ‘ π ;
f) π ππ2 π + πππ 2 π = π¦
π
2+
π₯
π
2 =
π¦2
π2 +
π₯2
π2 =
π¦2 + π₯2
π2 =
π2
π2 = 1 ;
g) 1 + π‘ππ2 π = 1 + π¦
π₯
2 = 1 +
π¦2
π₯2 =
π₯2 + π¦2
π₯2 =
π2
π₯2 = π ππ2 π ;
h) 1 + πππ‘ 2π = 1 + π₯
π¦
2 = 1 +
π₯2
π¦2 =
π¦ 2 + π₯2
π¦2 =
π2
π¦2 = πππ ππ2 π.
9. Prove that πππ ππ2 π΄ β π‘ππ2 π΄ β 1 = π‘ππ2 π΄.
Solution: πΏπ»π = 1 + πππ‘2π΄ β π‘ππ2 π΄ β 1
= π‘ππ2 π΄ + πππ‘2 π΄ β π‘ππ2 π΄ β 1
= π‘ππ2 π΄ + πππ‘ π΄ β π‘ππ π΄ 2 β 1
= π‘ππ2 π΄ + 1 β 1 = π‘ππ2 π΄ = π π»π.
10. Prove that πππ πΌ
π ππ πΌ+
π ππ πΌ
πππ ππ πΌ= 1.
Solution: πΏπ»π = πππ πΌ
1
πππ πΌ
+ π ππ πΌ
1
π ππ πΌ
= πππ 2 πΌ + π ππ2 πΌ = 1 = π π»π.
11. Prove that πππ 4 π΄ β π ππ4 π΄ = 2 πππ 2 π΄ β 1.
Solution: πΏπ»π = πππ 2 π΄ 2 β π ππ2 π΄ 2
= πππ 2 π΄ + π ππ2 π΄ πππ 2 π΄ β π ππ2 π΄
= 1 β πππ 2 π΄ β π ππ2 π΄
= πππ 2 π΄ β 1 β πππ 2 π΄
= πππ 2 π΄ β 1 + πππ 2 π΄ = 2πππ 2 π΄ β 1 = π π»π .
12. Prove that π ππ π β πππ‘ π 2 β πππ π β πππ ππ π 2 = 1.
Solution: πΏπ»π = 1
cos πβ
cos π
sin π
2β cos π β
1
sin π
2
= 1
sin π
2β
cos π
sin π
2 = cosec π 2 β cot π 2 = 1 = π π»π.
13. Prove that π ππ π β π‘ππ π β π ππ π = πππ π.
Solution: πΏπ»π = 1
πππ π β
π ππ π
πππ πβ π ππ π
= 1
πππ πβ
π ππ 2 π
πππ π=
1βπ ππ 2 π
πππ π =
πππ π 2
πππ π= πππ π = π π»π.
14. Prove that π‘ππ π + πππ‘ π = π ππ π β πππ ππ π.
Solution: πΏπ»π = π ππ π
πππ π +
πππ π
π ππ π
= π ππ 2 π + πππ 2 π
πππ πβ π ππ π
= 1
πππ π β
1
π ππ π = π ππ π β πππ ππ π = π π»π.
15. Prove that πππ π + π ππ π 2 + πππ π β π ππ π 2 = 2.
Solution:
πΏπ»π = πππ 2 π + π ππ2 π + 2 πππ π . π ππ π
+ πππ 2 π + π ππ2 π β 2 cos π . sin π = 1 + 1 = 2 = π π»π.
16. Prove that π‘ππ2πΌ + π‘ππ4πΌ = π ππ4πΌ β π ππ2πΌ.
Solution: πΏπ»π = π‘ππ2πΌ + π‘ππ4πΌ = π‘ππ2πΌ 1 + π‘ππ2πΌ
= π ππ2πΌ β 1 π ππ2πΌ = π ππ4πΌ β π ππ2πΌ = π π»π.
17. Prove that 1
1 β πππ πΌ+
1
1 + πππ πΌ= 2 πππ ππ2πΌ
Solution: πΏπ»π = 1 + πππ πΌ + 1 β πππ πΌ
1 β πππ πΌ 1 β πππ πΌ
=2
1 β πππ 2 πΌ =
2
π ππ 2 πΌ = 2 πππ ππ2πΌ = π π»π.
18. Prove that 1
1 + πππ 2 πΌ+
1
1 + π ππ 2 πΌ = 1.
Solution: πΏπ»π = 1
1 + πππ 2 πΌ +
1
1 + 1
πππ 2 πΌ
=1
1+ πππ 2 πΌ +
πππ 2 πΌ
1+ πππ 2 πΌ =
1 + πππ 2 πΌ
1 + πππ 2 πΌ= 1 = π π»π.
19. Prove that π ππ290Β° + πππ 260Β° β 2 π ππ245Β° +3
2π ππ230Β° =
5
8.
Solution: πΏπ»π = 12 + 1
2
2β 2
1
2
2+
3
2
1
2
2
= 1 + 1
4 β 1 +
3
8 =
1
4 +
3
8 =
5
8 = π π»π.
20. Prove that πππ‘2 π
6 β 2πππ 2 π
3 β
3
4π ππ2 π
4 β 4π ππ2 π
6 = 0.
Solution: πΏπ»π = 3 2β 2
1
2
2β
3
4 2
2+ 4
1
2
2
= 3 β 1
2 β
3
2 β 1 = 0 = π π»π.
21. Prove that πππ ππ245Β° β πππ 60Β° β sin90Β° β π ππ230Β° =4
3.
Solution: πΏπ»π = 2 2β
1
2β 1 β
2
3
2= 2 β
1
2β 1 β
4
3 =
4
3 = π π»π.
22. Prove that π‘ππ260Β° + 2 π‘ππ245Β° = 5.
Solution: πΏπ»π = 3 2
+ 2 = 3 + 2 = 5 = π π»π.
23. Prove that π‘ππ2 π
6+ π‘ππ2 π
4+ π‘ππ2 π
3=
13
3 .
Solution: πΏπ»π = 1
3
2+ 1 2 + 3
2
= 1
3 + 1 + 3 =
1 + 3 + 9
3 =
13
3= π π»π.
24. Prove that 1 β πππ 30Β°
1 + πππ 30Β° =
2 β 3
2 + 3
Solution: πΏπ»π =1 β
3
2
1 + 3
2
= 2 β 3
2
2 + 3
2
=2 β 3
2 + 3= π π»π.
25. Find π₯, π₯ π ππ2 π
4 . πππ 2 π
4= π‘ππ2 π
3
Solution: π₯ 1
2
2β
1
2
2= 3
2 βΉ
π₯
4= 3 β΄ π₯ = 12.
26. Prove that trigonometric ratios of 2ππ + π are same as the
trigonometric ratio of π.
Solution: π, 2π + π, 4π + π, ... are co-terminal angles. In
general π is co-terminal with the angle 2ππ + π, β π β π.
Since the points corresponding to the coterminal angles are the
same, the trigonometric ratios of coterminal angles are the
same. Thus π ππ (2ππ + π) = π ππ π, πππ (2ππ + π) = πππ π and
π‘ππ (2ππ + π) = π‘ππ π, β π β π.
27. Prove that trigonometric ratios of 2ππ β π, β π β π are
same as the trigonometric ratio of β π.
Solution: β π, 2π β π, 4π β π, ... are co-terminal angles. In
general β π is co-terminal with the angle 2ππ β π, β π β π.
Since the points corresponding to the coterminal angles are the
same, the trigonometric ratios of coterminal angles are the
same. Thus π ππ (2ππ β π) = π ππ (β π), πππ (2ππ β π) = πππ ( β π)
and π‘ππ (2ππ β π) = π‘ππ ( β π), β π β π.
28. Prove that π ππ (180Β° β π) = π ππ π.
Solution: π ππ (180Β° β π) = π ππ 180Β°. πππ π β πππ 180Β°. π ππ π
= 0. πππ π β (β 1). π ππ π = π ππ π.
29. Prove that πππ (180Β° β π) = β πππ π.
Solution: πππ (180Β° β π) = πππ 180Β°. πππ π + π ππ 180Β°. π ππ π
= (β 1). πππ π + 0. π ππ π = β πππ π.
30. Prove that π‘ππ (180Β° β π) = β π‘ππ π.
Solution: π‘ππ (180Β° β π) = π‘ππ 180Β° β π‘ππ π
1 + π‘ππ 180Β°.π‘ππ π
= 0 β π‘ππ π
1 + 0.π‘ππ π = β π‘ππ π.
31. Prove that π ππ (180Β° + π) = β π ππ π.
Solution: π ππ (180Β° + π) = π ππ 180Β°. πππ π + πππ 180Β°. π ππ π
= 0. πππ π + (β 1). π ππ π = β π ππ π.
32. Prove that πππ (180Β° + π) = β πππ π.
Solution: πππ (180Β° + π) = πππ 180Β°. πππ π β π ππ 180Β°. π ππ π
= (β 1). πππ π β 0. π ππ π = β πππ π.
33. Prove that π‘ππ (180Β° + π) = π‘ππ π.
Solution: π‘ππ (180Β° + π) = π‘ππ 180Β° + π‘ππ π
1 β π‘ππ 180Β°.π‘ππ π
= 0 + π‘ππ π
1 β 0.π‘ππ π = π‘ππ π.
34. Prove that π ππ (270Β° β π) = β πππ π.
Solution: π ππ (270Β° β π) = π ππ 270Β°. πππ π β πππ 270Β°. π ππ π
= β 1. πππ π β 0. π ππ π = β πππ π.
35. Prove that πππ (270Β° β π) = π ππ π.
Solution: πππ (270Β° β π) = πππ 270Β°. πππ π + π ππ 270Β°. π ππ π
= 0. πππ π + (β 1). π ππ π = β π ππ π.
36. Prove that π‘ππ (270Β° β π) = πππ‘ π.
Solution: π‘ππ (270Β° β π) = π‘ππ (180Β° + 90Β° β π)
= π‘ππ ( 90Β° β π) = πππ‘ π.
37. Prove that π ππ (270Β° + π) = β πππ π.
Solution: π ππ (270Β° + π) = π ππ 270Β°. πππ π + πππ 270Β°. π ππ π
= β 1. πππ π + 0. π ππ π = β πππ π.
38. Prove that πππ (270Β° + π) = π ππ π.
Solution: πππ (270Β° + π) = πππ 270Β°. πππ π β π ππ 270Β°. π ππ π
= 0. πππ π β (β 1). π ππ π = π ππ π.
39. Prove that π‘ππ (270Β° + π) = β πππ‘ π.
Solution: π‘ππ (270Β° + π) = π‘ππ (180Β° + 90Β° + π)
= π‘ππ ( 90Β° + π) = β πππ‘ π.
40. Prove that π‘ππ (π₯ + π¦) = π‘ππ π₯ + π‘ππ π¦
1 β π‘ππ π₯ .π‘ππ π¦
Solution: π‘ππ (π₯ + π¦) = π ππ (π₯ + π¦)
πππ (π₯ + π¦) =
π ππ π₯ .πππ π¦ + πππ π₯ .π ππ π¦
πππ π₯ .πππ π¦ β π ππ π₯ .π ππ π¦.
Dividing the numerator and the denominator by πππ π₯. πππ π¦ we
get, π‘ππ (π₯ + π¦) =
π ππ π₯ .πππ π¦
πππ π₯ .πππ π¦ +
πππ π₯ .π ππ π¦
πππ π₯ .πππ π¦
πππ π₯ .πππ π¦
πππ π₯ .πππ π¦ β
π ππ π₯ .π ππ π¦
πππ π₯ .πππ π¦
= π‘ππ π₯ + π‘ππ π¦
1 β π‘ππ π₯ .π‘ππ π¦ .
41. Prove that π‘ππ (π₯ β π¦) = π‘ππ π₯ β π‘ππ π¦
1 + π‘ππ π₯ .π‘ππ π¦
Solution: π‘ππ (π₯ β π¦) = π ππ (π₯ β π¦)
πππ (π₯ β π¦) =
π ππ π₯ .πππ π¦ β πππ π₯ .π ππ π¦
πππ π₯ .πππ π¦ + π ππ π₯ .π ππ π¦.
Dividing the numerator and the denominator by πππ π₯. πππ π¦ we
get, π‘ππ (π₯ β π¦) =
π ππ π₯ .πππ π¦
πππ π₯ .πππ π¦ β
πππ π₯ .π ππ π¦
πππ π₯ .πππ π¦
πππ π₯ .πππ π¦
πππ π₯ .πππ π¦ +
π ππ π₯ .π ππ π¦
πππ π₯ .πππ π¦
= π‘ππ π₯ β π‘ππ π¦
1 + π‘ππ π₯ .π‘ππ π¦ .
42. Prove that πππ‘ (π₯ + π¦) = πππ‘ π₯ .πππ‘ π¦ β 1
πππ‘ π₯ + πππ‘ π¦
Solution: We know that π‘ππ (π₯ + π¦) =π‘ππ π₯ + π‘ππ π¦
1 β π‘ππ π₯ .π‘ππ π¦ .
β΄ πππ‘ (π₯ + π¦) =1 β π‘ππ π₯ .π‘ππ π¦
π‘ππ π₯ + π‘ππ π¦
=1 β
1
πππ‘ π₯ .πππ‘ π¦1
πππ‘ π₯ +
1
πππ‘ π¦
=
πππ‘ π₯ .πππ‘ π¦ β 1
πππ‘ π₯ .πππ‘ π¦πππ‘ π¦ + πππ‘ π₯
πππ‘ π₯ .πππ‘ π¦
= πππ‘ π₯ .πππ‘ π¦ β 1
πππ‘ π₯ + πππ‘ π¦
43. Prove that πππ‘ (π₯ β π¦) = πππ‘ π₯ .πππ‘ π¦ + 1
πππ‘ π¦ β πππ‘ π₯.
Solution: We know that π‘ππ (π₯ β π¦) =π‘ππ π₯ β π‘ππ π¦
1 + π‘ππ π₯ .π‘ππ π¦ .
β΄ πππ‘ (π₯ β π¦) =1 + π‘ππ π₯ .π‘ππ π¦
π‘ππ π₯ β π‘ππ π¦
= 1 +
1
πππ‘ π₯ .πππ‘ π¦1
πππ‘ π₯ β
1
πππ‘ π¦
=
πππ‘ π₯ .πππ‘ π¦ + 1
πππ‘ π₯ .πππ‘ π¦πππ‘ π¦ β πππ‘ π₯
πππ‘ π₯ .πππ‘ π¦
=πππ‘ π₯ .πππ‘ π¦ + 1
πππ‘ π¦ β πππ‘ π₯.
44. Prove that π ππ (π₯ + π¦). π ππ (π₯ β π¦) = π ππ2 π₯ β π ππ2 π¦
Solution: π ππ (π₯ + π¦). π ππ (π₯ β π¦)
= (π ππ π₯. πππ π¦ + πππ π₯. π ππ π¦). (π ππ π₯. πππ π¦ β πππ π₯. π ππ π¦)
= π ππ2 π₯. πππ 2π¦ β πππ 2 π₯. π ππ2 π¦
= π ππ2 π₯(1 β π ππ2π¦) β (1 β π ππ2 π₯)π ππ2 π¦
= π ππ2 π₯ β π ππ2 π₯. π ππ2 π¦ β π ππ2 π¦ + π ππ2 π₯. π ππ2 π¦
= π ππ2 π₯ β π ππ2 π¦.
45. Prove that πππ (π₯ + π¦). πππ (π₯ β π¦) = πππ 2 π₯ β π ππ2 π¦
Solution: πππ (π₯ + π¦). πππ (π₯ β π¦)
= (πππ π₯. πππ π¦ β π ππ π₯. π ππ π¦).(πππ π₯. πππ π¦ + π ππ π₯. π ππ π¦)
= πππ 2 π₯. πππ 2π¦ β π ππ2 π₯. π ππ2 π¦
= πππ 2 π₯(1 β π ππ2π¦) β (1 β πππ 2 π₯)π ππ2 π¦
= πππ 2 π₯ β πππ 2 π₯. π ππ2 π¦ β π ππ2 π¦ + πππ 2 π₯. π ππ2 π¦
= πππ 2 π₯ β π ππ2 π¦.
46. Find the values of the following
a) π ππ 15Β° ; b) πππ 15Β° ; c) π‘ππ 15Β° ;
d) π ππ 75Β° ; e) πππ 75Β° ; f) π‘ππ 75Β°.
Solution: a) π ππ 15Β° = π ππ (45Β° β 30Β°)
= π ππ 45Β°. πππ 30Β° β πππ 45Β°. π ππ 30Β°
= 1
2. 3
2 β
1
2.
1
2 =
3 β 1
2 2.
b) πππ 15Β° = πππ (45Β° β 30Β°)
= πππ 45Β°. πππ 30Β° + π ππ 45Β°. π ππ 30Β°
= 1
2. 3
2 +
1
2.
1
2 =
3 + 1
2 2
c) π‘ππ 15Β° = π‘ππ (45Β° β 30Β°)
= π‘ππ 45Β° β π‘ππ 30Β°
1 + π‘ππ 45Β°.π‘ππ 30Β° =
1 β 1
3
1 + 1.1
3
= 3 β 1
3 + 1
= 3 β 1
3 + 1. 3 β 1
3 β 1 =
3 β 1 2
32
β 1 =
3 + 1 β 2 3
2
= 4 β 2 3
2 = 2 β 3.
d) π ππ 75Β° = π ππ (45Β° + 30Β°)
= π ππ 45Β°. πππ 30Β° + πππ 45Β°. π ππ 30Β°
= 1
2. 3
2 +
1
2.
1
2 =
3 + 1
2 2.
e) πππ 75Β° = πππ (45Β° + 30Β°)
= πππ 45Β°. πππ 30Β° β π ππ 45Β°. π ππ 30Β°
= 1
2. 3
2 β
1
2.
1
2 =
3 β 1
2 2
f) π‘ππ 75Β° = π‘ππ (45Β° + 30Β°)
= π‘ππ 45Β° + π‘ππ 30Β°
1 β π‘ππ 45Β°.π‘ππ 30Β° =
1 + 1
3
1 β 1.1
3
= 3 + 1
3 β 1
= 3 + 1
3 β 1. 3 + 1
3 + 1 =
3 + 1 2
32
β 1 =
3 + 1 + 2 3
2
= 4 + 2 3
2 = 2 + 3.
47. Find the values of the following
a) πππ ππ 15Β° ; b) π ππ 15Β° ; c) πππ‘ 15Β°;
d) πππ ππ 75Β° ; e) π ππ 75Β° ; f) πππ‘ 75Β°.
Solution: a) πππ ππ 15Β° = 1
π ππ 15Β° =
2 2
3 β 1 =
2 2
3 β 1. 3 + 1
3 + 1
= 2 2( 3 + 1)
3 β 1 = 2( 3 + 1) = 6 + 2.
b) π ππ 15Β° = 1
πππ 15Β° =
2 2
3 + 1 =
2 2
3 + 1. 3 β 1
3 β 1 =
2 2( 3 β 1)
3 β 1
= 2( 3 β 1) = 6 β 2.
c) πππ‘ 15Β° = 1
π‘ππ 15Β° =
1
2 β 3 =
1
2 β 3.
2 + 3
2 + 3 =
2 + 3
4 β 3 = 2 + 3.
d) πππ ππ 75Β° = 1
π ππ 75Β° =
2 2
3 + 1 =
2 2
3 + 1. 3 β 1
3 β 1 =
2 2( 3 β 1)
3 β 1
= 2( 3 β 1) = 6 β 2.
e) π ππ 75Β° = 1
πππ 75Β° =
2 2
3 β 1 =
2 2
3 β 1. 3 + 1
3 + 1 =
2 2( 3 + 1)
3 β 1
= 2( 3 + 1) = 6 + 2.
f) πππ‘ 75Β° = 1
π‘ππ 75Β° =
1
2 + 3 =
1
2 + 3.
2 β 3
2 β 3 =
2 + 3
4 β 3 = 2 + 3.
48. Show that
πππ 45Β° β π . πππ 45Β° + π β π ππ 45Β° β π . π ππ 45Β° + π = 0
Solution:
πΏπ»π = πππ 45Β° β π . πππ 45Β° + π β π ππ 45Β° β π . π ππ 45Β° + π
= πππ 45Β° β π + 45Β° + π
= πππ 45Β° β π + 45Β° + π = cos 90Β° = 0.
49. Show that π‘ππ π
4+ π β π‘ππ
π
4β π = 1.
Solution: πΏπ»π = π‘ππ π
4 + π . π‘ππ
π
4β π
= π‘ππ
π
4 + π‘πππ
1 β π‘ππ π
4.π‘πππ
π‘ππ
π
4 β π‘πππ
1 + π‘ππ π
4.π‘πππ
= 1 + π‘πππ
1 β π‘πππ
1 β π‘πππ
1 + π‘πππ = 1 = RHS.
50. Show that π ππ π₯ + π¦ + π ππ π₯ + π¦
πππ π₯ + π¦ + πππ π₯ + π¦ = π‘ππ π₯.
Solution: LHS =π ππ π₯ + π¦ + π ππ π₯ β π¦
πππ π₯ + π¦ + πππ π₯ β π¦
=π πππ₯ πππ π¦ + πππ π₯ π πππ¦ + π πππ₯ πππ π¦ β πππ π₯ π πππ¦
πππ π₯ πππ π¦ β π πππ₯ π πππ¦ + πππ π₯ πππ π¦ + π πππ₯ π πππ¦ =
2π πππ₯ πππ π¦
2πππ π₯ πππ π¦= tan π₯.
51. If π‘πππ΄ = 1
2 , π‘πππ΅ =
1
3 show that π΄ + π΅ =
π
4
Solution: π‘ππ π΄ + π΅ =π‘ππ π΄ + π‘ππ π΅
1 β π‘ππ π΄ π‘ππ π΅ =
1
2 +
1
3
1 β 1
2 Γ
1
3
= 5 6
1 β 1 6 =
5 6
5 6 = 1 = tan
Ο
4 β΄ π΄ + π΅ =
π
4
52. Prove that π‘ππ 85Β° β π‘ππ 25Β° = 3(1 + π‘ππ 85Β° β π‘ππ 25Β°
Solution:
π‘ππ 60Β° = 3 βΉ π‘ππ 85Β° β 25Β° = 3
βΉπ‘ππ 85Β° β π‘ππ 25Β°
1 + π‘ππ 85Β°βπ‘ππ 25Β° = 3
βΉ π‘ππ 85Β° β π‘ππ 25Β° = 3(1 + π‘ππ 85Β° β π‘ππ 25Β°)
53. Prove that a) π‘ππ 69Β° + π‘ππ 66Β°
1 β π‘ππ 69Β°β π‘ππ 66Β° = β 1 ; b)
πππ 17Β° + π ππ17Β°
πππ 17Β°β π ππ17Β°= π‘ππ62Β°
Solution:
a) πΏπ»π =π‘ππ 69Β° + π‘ππ 66Β°
1 β π‘ππ 69Β°β π‘ππ 66Β° = π‘ππ 69Β° + 66Β° = π‘ππ 135Β°
= π‘ππ 90Β° + 45Β° = β πππ‘ 45Β° = β 1 = π π»π .
b) Dividing the numerator and denominator by πππ 17Β°, we get,
πΏπ»π =πππ 17Β° + π ππ 17Β°
πππ 17Β° β π ππ 17Β° =
1 + π‘π π17Β°
1 β π‘ππ 17Β°
= π‘ππ 45Β° + π‘ππ 17Β°
1 β π‘ππ 45Β°β π‘ππ 17Β° = π‘ππ 45Β° + 17Β° = π‘ππ 62Β° = π π»π.
54. Prove that π‘ππ 3π΄ β π‘ππ 2π΄ β π‘ππ π΄ = π‘ππ π΄ β π‘ππ 2π΄ β π‘ππ 3π΄
Solution: π‘ππ 3π΄ = π‘ππ π΄ + 2π΄ =π‘ππ π΄ + π‘ππ 2π΄
1 β π‘ππ π΄ π‘ππ 2π΄
βΉ π‘ππ 3π΄ 1 β π‘ππ π΄ π‘ππ 2π΄ = π‘ππ π΄ + π‘ππ 2π΄
βΉ π‘ππ 3π΄ β π‘ππ π΄. π‘ππ 2π΄. π‘ππ 3π΄ = π‘ππ π΄ + π‘ππ 2π΄
βΉ π‘ππ 3π΄ β π‘ππ 2π΄ β π‘ππ π΄ = π‘ππ π΄ β π‘ππ 2π΄ β π‘ππ 3π΄.
55. Prove that π ππ 2π΄ = 2π ππ π΄. πππ π΄.
Solution: We know that π ππ (π₯ + π¦) = π ππ π₯. πππ π¦ + πππ π₯. π ππ π¦.
By taking π₯ = π¦ = π΄, we get,
π ππ (π΄ + π΄) = π ππ π΄. πππ π΄ + πππ π΄. π ππ π΄
β π ππ 2π΄ = 2π ππ π΄. πππ π΄.
56. Prove that πππ 2π΄ = πππ 2π΄ β π ππ2 π΄.
Solution: We know that πππ (π₯ + π¦) = πππ π₯. πππ π¦ β π ππ π₯. π ππ π¦.
By taking π₯ = π¦ = π΄, we get,
πππ (π΄ + π΄) = πππ π΄. πππ π΄ β π ππ π΄. π ππ π΄
β πππ 2π΄ = πππ 2π΄ β π ππ2π΄.
57. Prove that πππ 2π΄ = 2. πππ 2π΄ β 1
Solution: We know that πππ (π₯ + π¦) = πππ π₯. πππ π¦ β π ππ π₯. π ππ π¦.
By taking π₯ = π¦ = π΄, we get,
πππ (π΄ + π΄) = πππ π΄. πππ π΄ β π ππ π΄. π ππ π΄
β πππ 2π΄ = πππ 2π΄ β π ππ2π΄
= πππ 2π΄ β (1 β πππ 2π΄)
= πππ 2π΄ β 1 + πππ 2π΄ = πππ 2π΄ β 1.
58. Prove that πππ 2π΄ = 1 β 2. π ππ2π΄
Solution: We know that πππ (π₯ + π¦) = πππ π₯. πππ π¦ β π ππ π₯. π ππ π¦.
By taking π₯ = π¦ = π΄, we get,
πππ (π΄ + π΄) = πππ π΄. πππ π΄ β π ππ π΄. π ππ π΄
β πππ 2π΄ = πππ 2π΄ β π ππ2π΄
= (1 β π ππ2π΄) β π ππ2π΄ = 1 β 2π ππ2π΄.
59. Prove that π‘ππ 2π΄ = 2π‘ππ π΄
1 β π‘ππ 2π΄.
Solution: We know that π‘ππ (π₯ + π¦) = π‘ππ π₯ + π‘ππ π¦
1 β π‘ππ π₯ .π‘ππ π¦ .
By taking π₯ = π¦ = π΄, we get, π‘ππ (π΄ + π΄) = π‘ππ π΄ + π‘ππ π΄
1 β π‘ππ π΄.π‘ππ π΄
β π‘ππ 2π΄ = 2π‘ππ π΄
1 β π‘ππ 2π΄.
60. Prove that π ππ 2π΄ = 2π‘ππ π΄
1 + π‘ππ 2π΄.
Solution: 2.π‘ππ π΄
1 + π‘ππ 2 π΄ =
2.π‘ππ π΄
π ππ 2 π΄ = 2. π‘ππ π΄. πππ 2 π΄ = 2
π ππ π΄
πππ π΄πππ 2 π΄
= 2. π ππ π΄. πππ π΄ = π ππ 2π΄.
61. Prove that πππ 2π΄ = 1 β π‘ππ 2 π΄
1 + π‘ππ 2 π΄.
Solution:
1 β π‘ππ 2 π΄
1 + π‘ππ 2 π΄ =
1 β π‘ππ 2 π΄
π ππ 2 π΄ =
1
π ππ 2 π΄ β
π‘ππ 2 π΄
π ππ 2 π΄ = πππ 2 π΄ β π‘ππ2 π΄. πππ 2 π΄
= πππ 2 π΄ β π ππ2 π΄
πππ 2 π΄πππ 2 π΄ = πππ 2 π΄ β π ππ2 π΄ = πππ 2π΄.
62. Prove that π‘ππ2 π΄ = 1 β πππ 2π΄
1 + πππ 2π΄.
Solution:
We know that πππ 2π΄ = 1 β 2. π ππ2π΄. β΄ 2. π ππ2π΄ = 1 β πππ 2π΄ ;
Also πππ 2π΄ = 2. πππ 2π΄ β 1 β΄ 2. πππ 2π΄ = 1 + πππ 2π΄.
β΄ 1 β πππ 2π΄
1 + πππ 2π΄ =
2.π ππ 2π΄
2.πππ 2π΄ = π‘ππ2π΄.
63. Find the values of a. π ππ 221
2Β° ; b. πππ 22
1
2Β° ; c. π‘ππ 22
1
2Β°.
Solution:
a. We know that 2. π ππ2 π
2 = 1 β πππ π.
By taking π = 45Β° we get
2. π ππ2 221
2Β° = 1 β πππ 45Β° = 1 β
1
2 =
2 β 1
2.
β΄ π ππ2 221
2Β° =
2 β 1
2 2 =
2( 2 β 1)
2 2. 2 =
2 β 2
4
β΄ π ππ 221
2Β° = 2 β 2
4 =
2 β 2
2
b. We know that 2. πππ 2 π
2 = 1 + πππ π.
By taking π = 45Β° we get
2. πππ 2 221
2Β° = 1 + πππ 45Β° = 1 +
1
2 =
2 + 1
2.
β΄ πππ 2 221
2Β° =
2 + 1
2 2 =
2( 2 + 1)
2 2. 2 =
2 + 2
4.
β΄ πππ 221
2Β° =
2 + 2
4 =
2 + 2
2.
c. We know that, π‘ππ2 π
2 =
1 β πππ π
1 + πππ π .
By taking π = 45Β° we get,
π‘ππ2 221
2Β° =
1 β πππ 45Β°
1 + πππ 45Β° =
1 β 1 2
1 + 1 2 =
2 β 1
2 + 1
= 2 β 1
2 + 1. 2 β 1
2 β 1 =
2 β 1 2
2 β 1 = 2 β 1
2.
β΄ π‘ππ 221
2Β° = 2 β 1.
64. If π ππ π΄ =3
8 and π΄ is acute angle, find π ππ 2π΄.
Solution: Given π ππ π΄ =3
8.
Since π΄ is acute,
πππ π΄ = 1 β π ππ2π΄ = 1 β 9
64 =
64 β 9
64 =
55
8.
β΄ π ππ 2π΄ = 2 π ππ π΄. πππ π΄ = 2 β3
8β 55
8 =
3 55
32.
65. Prove that πππ 4π΄ β π ππ4π΄ = cos 2π΄.
Solution: πΏπ»π = πππ 2π΄ 2 β π ππ2π΄ 2
= πππ 2π΄ + π ππ2π΄ πππ 2π΄ β π ππ2π΄
= 1 β cos 2π΄ = cos 2π΄ = π π»π.
66. Prove that 2 + 2 + 2 πππ 4π = 2. πππ π
Solution: πΏπ»π = 2 + 2(1 + πππ 4π) = 2 + 2.2πππ 2 2π
= 2 + 2 πππ 2π = 2(1 + πππ 2π)
= 2 β 2πππ 2 2π = 2πππ π = π π»π.
67. Prove that 1 + π‘ππ 2(45Β° β π)
1 β π‘ππ 2(45Β° β π) = πππ ππ 2π.
Solution: 1 + π‘ππ 2(45Β° β π)
1 β π‘ππ 2(45Β° β π) =
1 + π‘ππ 2π‘
1 β π‘ππ 2π‘ =
1
πππ 2π‘ where π‘ = 45Β° β π.
= π ππ 2π‘ = π ππ 2 45Β° β π = π ππ 90Β° β 2π = πππ ππ 2π.
68. Prove that 8 πππ 3 π
9 β 6 πππ
π
9 = 1.
Solution: 8 πππ 3 π
9β 6 πππ
π
9 = 2 4 πππ 3
π
9 β 3 πππ
π
9
= 2 πππ 3 βπ
9 = 2 πππ
π
3 = 2 β
1
2 = 1.
69. Prove that 1 + π ππ π β πππ π
1 + π ππ π + πππ π = π‘ππ
π
2
Solution: 1 + π ππ π β πππ π
1 + π ππ π + πππ π =
1 β πππ π + π ππ π
1 + πππ π + π ππ π
=2π ππ 2
π
2 + 2 π ππ π
2 βπππ
π
2
2πππ 2 π
2 + 2 π ππ π
2 βπππ
π
2
=2 π ππ π
2 πππ
π2 + π ππ
π2
2 πππ π
2 πππ
π
2 + π ππ
π
2
= π‘ππ π
2.
70. Prove that πππ 10Β° β πππ 50Β° = π ππ 20Β°
Solution: πΏπ»π = β 2 π ππ 10 Β° + 50 Β°
2 π ππ
10Β° β 50Β°
2
= β 2 π ππ 30Β° π ππ β20Β° = β (β 2 β1
2π ππ 20Β°) = π ππ20Β°.
71. Find the general solution of π ππ π =1
2
Solution: π ππ π =1
2= π ππ
π
6. Principal angle πΌ =
π
6.
β΄ the general solution is π₯ = ππ + (β1)π β π
6 ,π β π.
72. Find the general solution of π ππ π = β 2.
Solution: π ππ π = β 2 βΉ cos π = β1
2.
Principal angle πΌ = π βπ
4 =
3π
4
β΄ general solution is π = 2ππ Β± 3π
4 , π β π.
73. Find the general solution of π ππ 2π₯ = 4 πππ π₯.
Solution: π ππ 2π₯ = 4 πππ π₯ βΉ 2π ππ π₯ . πππ π₯ = 4 πππ π₯
βΉ 2π ππ π₯. πππ π₯ β 4πππ π₯ = 0 βΉ 2πππ π₯(π ππ π₯ β 2) = 0
βΉ πππ π₯ = 0 ππ π ππ π₯ = 2.
Since π ππ π₯ = 2 > 1, the equation π ππ π₯ = 2 has no solution.
πππ π₯ = 0 βΉ π₯ = 2π + 1 π
2 , π β Z.
This is the general solution of the given equation.
Three mark questions
1. The angles of a triangle are in the ratio 4: 5: 6. Find them in
radian and degree.
Solution: Let A, B and C be the angles of the triangle.
Now π΄: π΅: πΆ = 4: 5: 6 βΉ π΄ = 4π, π΅ = 5π,πΆ = 6π
π΄ + π΅ + πΆ = π βΉ 4π + 5π + 6π = π βΉ 15π = π β΄ π =π
15
β΄ π΄ = 4π = 4 Γπ
15=
4π
15 , π΅ = 5π = 5 Γ
π
15=
π
3, πΆ = 6π = 6 Γ
π
15=
2π
3.
π΄ + π΅ + πΆ = 180Β° βΉ 15π = 180Β° β΄ π =180Β°
15= 12Β°
β΄ π΄ = 4π = 4 Γ 12Β° = 48Β°, π΅ = 5 Γ 12Β° = 60Β°, πΆ = 6π = 6 Γ 12Β° = 72Β°
2. Define the six trigonometric functions.
Solution: Consider a circle of radius π centered at the origin
of the Cartesian coordinate system. Let
π(π₯, π¦) be a point on the circle. Join ππ.
Define angle π having ππ has the initial
direction and ππ as the terminal direction.
Then for any position of the point π(π₯, π¦)
on the circle,
i. sine of the angle π is denoted by π ππ π and is defined by
π ππ π = π¦
π.
ii. cosine of the angle π is denoted by πππ π and is defined by
πππ π = π₯
π.
iii. tangent of the angle π is denoted by π‘ππ π and is defined by
π‘ππ π = π¦
π₯.
iv. cotangent of the angle π is denoted by πππ‘ π and is defined
by πππ‘ π = π₯
π¦.
v. secant of the angle π is denoted by π ππ π and is defined by
π ππ π = π
π₯.
vi. cosecant of the angle π is denoted by πππ ππ π or ππ π π and
is defined by πππ ππ π = π
π¦.
3. If sin π =3
5 and π is acute, find all the other trigonometric
ratios.
Solution: Given sin π = 3
5=
π΄π΅
ππ΄.
Therefore π΄π΅ = 3 and ππ΄ = 5
Now ππ΅2 = ππ΄2 β π΄π΅2 = 25 β 9 = 16.
Therefore ππ΅ = 4.
We have, cosπ = 1 β π ππ2π = 1 β 9
25 =
25 β 9
25 =
16
25=
4
5 ;
tan π =sin π
cos π=
35
45
=3
4 ;
cot π =cos π
sin π=
45
35
=4
3 ;
π πππ =1
cos π=
14
5 =
5
4 ;
πππ πππ =1
sin π=
13
5 =
5
3 .
4. Prove that 1 β π ππ π΄ + πππ π΄ 2 = 2 1 β π ππ π΄ 1 + πππ π΄
Solution:
πΏπ»π = 1 β π ππ π΄ + πππ π΄ 2
= 1 β π ππ π΄ 2 + πππ 2 π΄ + 2 1 β π ππ π΄ β πππ π΄
= 1 β π ππ π΄ 2 + 1 β π ππ2 π΄ + 2 1 β π ππ π΄ β πππ π΄
= 1 β π ππ π΄ 2 + 1 β π ππ π΄ β 1 + π ππ π΄ + 2 1 β π ππ π΄ β πππ π΄
= 1 β π ππ π΄ 1 β π ππ π΄ + 1 + π ππ π΄ + 2 πππ π΄
= 1 β π ππ π΄ 2 + 2 πππ π΄ = 2 1 β π ππ π΄ 1 + πππ π΄ = π π»π.
5. Find the trigonometric ratios of the quadrantal angles
a) 0Β° ; b) 90Β° π
2 ; c) 180Β° (π) ; d) 270Β°
3π
2 and e) 360Β° (2π).
Solution: Consider the points, π΄(1, 0), π΅(0, 1), πΆ(β 1, 0) and
π·(0, β1) corresponding to the quadrantal angles π
2 (90Β°),
π (180Β°),3π
2 (270Β°) and 2π (360Β°),
lying on a circle of unit radius
centered at the origin of a Cartesian
co-ordinate system. Let π(π, π) be
the point on the circle such that
angle πππ = π₯.
Then π ππ π₯ = π and πππ π₯ = π.
a)When π(π, π) coincides with the point π΄(1, 0), the angle π₯
becomes 0Β°.
β΄ πππ 0 = 1 and π ππ 0 = 0.
β΄ π‘ππ 0 = 0
1 = 0, πππ‘ 0 =
1
0 = β, πππ ππ 0 =
1
0 = β, π ππ 0 =
1
1 = 1.
b) When π(π, π) coincides with the point π΅(0, 1), the angle π₯
becomes 90Β°.
β΄ πππ 90Β° = 0 and π ππ 90Β° = 1.
β΄ π‘ππ 90Β° = 1
0 = β, πππ‘ 90Β° =
0
1 = 0,
πππ ππ 90Β° = 1
1 = 1, π ππ 90Β° =
1
0 = β.
c) When π(π, π) coincides with the point πΆ(β 1, 0), the angle π₯
becomes 180Β°.
β΄ πππ 180Β° = β 1 and π ππ 180Β° = 0.
β΄ π‘ππ 180Β° = 0
β 1 = 0, πππ‘ 180Β° =
β 1
0 = β,
πππ ππ 180Β° = 1
0 = β, π ππ 180Β° =
1
β 1 = β 1.
d) When π(π, π) coincides with the point π·(0, β1), the angle π₯
becomes 270Β°.
β΄ πππ 270Β° = 0 and π ππ 270Β° = β 1.
β΄ π‘ππ 270Β° = β 1
0 = β, πππ‘ 270Β° =
0
β 1 = 0,
πππ ππ 270Β° = 1
β 1 = β 1, π ππ 270Β° =
1
0 = β.
e) When π(π, π) coincides with the point π΄(1, 0), after one
complete rotation, the angle π₯ becomes 360Β°.
β΄ πππ 360Β° = 1 and π ππ 360Β° = 0.
β΄ π‘ππ 360Β° = 0
1 = 0, πππ‘ 360Β° =
1
0 = β,
πππ ππ 360Β° = 1
0 = β, π ππ 360Β° =
1
1 = 1.
6. Find the trigonometric ratios of a) 45Β°; b) 30Β° and 60Β°.
Solution:
a) The trigonometric ratios of 45Β°.
Consider a right angled triangle π΄π΅πΆ, right angled
at π΅. Now angle π΄πΆπ΅ = 45Β°.
Let π΄π΅ = π₯. Then π΅πΆ = π₯.
Now π΄πΆ2 = π΄π΅2 + π΅πΆ2 = π₯2 + π₯2 = 2π₯2.
β΄ π΄πΆ = π₯ 2. Now,
π ππ 45Β° = π΄π΅
π΄πΆ=
π₯
π₯ 2=
1
2 ;
πππ 45Β° = π΅πΆ
π΄πΆ=
π₯
π₯ 2=
1
2 ;
π‘ππ 45Β° = π΄π΅
π΅πΆ=
π₯
π₯= 1.
Similarly, πππ‘ 45Β° = 1, πππ ππ 45Β° = 2, π ππ 45Β° = 2.
b) The trigonometric ratios of 30Β° and 60Β°.
Consider an equilateral triangle π΄π΅πΆ. Draw π΄π·
perpendicular to π΅πΆ. Then π· is the midpoint of
π΅πΆ. Let π΅π· = π₯. Then π΄π΅ = 2π₯.
Now π΄π·2 = π΄π΅2 β π΅π·2 = (2π₯)2 β π₯2 = 4π₯2 β
π₯2 = 3π₯2. β΄ π΄π· = π₯ 3.
Also, β π΄π΅πΆ = 60Β° and β π΅π΄π· = 30Β°.
π ππ 60Β° = π΄π·
π΄π΅=
π₯ 3
2π₯=
3
2 ;
πππ 60Β° = π΅π·
π΄π΅=
π₯
2π₯=
1
2 ;
π‘ππ 60Β° = π΄π·
π΅π·=
π₯ 3
π₯= 3.
Similarly, πππ‘ 60Β° = 1
3, πππ ππ 60Β° =
2
3, π ππ 60Β° = 2.
π ππ 30Β° = π΅π·
π΄π΅ =
π₯
2π₯ =
1
2 ;
πππ 30Β° = π΄π·
π΄π΅ =
π₯ 3
2π₯ =
3
2 ;
π‘ππ 30Β° = π΅π·
π΄π· =
π₯
π₯ 3 =
1
3.
Similarly, πππ‘ 30Β° = 3, πππ ππ 30Β° = 2, π ππ 30Β° = 2
3.
7. Prove that π‘ππ 260Β° β 2 π‘ππ 245Β°
3 π ππ 245Β° β π ππ90Β° + πππ 260Β° β πππ 2 0Β° =
4
7
Solution: πΏπ»π = 3
2 β 2 1 2
3 1
2
2β1 +
1
2
2β 1 2
=3 β 23
2 +
1
4
= 17
4
= 4
7 = π π»π.
8. Find π₯, π₯ sin 30Β° β πππ 245Β° =πππ‘ 230Β°β π ππ60Β°βπ‘ππ 45Β°
πππ ππ 245Β° β πππ ππ 30Β°
Solution: π₯ β1
2β
1
2
2=
3 2β 2 β 1
2 2β 2
βΉ π₯ β1
4 =
6
4. β΄ π₯ = 6.
9. Find π₯, π₯ π ππ230Β° + π ππ245Β° + π ππ260Β° β πππ 230Β° = 0
Solution: π₯ 1
2
2+
1
2
2+
3
2
2
β 3
2
2
= 0
βΉ π₯ 1
4 +
1
2 +
3
4 β
3
4= 0 βΉ π₯
1 + 2 + 3
4 =
3
4 β΄ π₯ =
3
6 =
1
2 .
10. Find the domain of π(π₯) = π ππ π₯ and π(π₯) = πππ ππ π₯.
Solution: Recall the definition, π ππ π = π¦
π.
This is defined for any π¦ β π .
β΄ domain of the function π¦ = π ππ π₯ is the set of all reals, π .
We know that π ππ 0 = 0, π ππ π = 0 and π ππ 2π = 0.
By using π ππ (2ππ + π) = π ππ π we get
π ππ (2ππ + π) = 0, β π β π.
This gives π ππ ππ = 0, β π β π.
Thus π ππ π₯ = 0 when π₯ = ππ, π β π.
Since πππ ππ π₯ = 1
π ππ π₯, the function π¦ = πππ ππ π₯ is defined only
when π ππ π₯ β 0. Thus the domain of the function,
π¦ = πππ ππ π₯, is {π₯ βΆ π₯ β ππ, π β π }.
11. Find the domain of π(π₯) = πππ π₯ and π(π₯) = π ππ π₯.
Solution: Recall the definition, πππ π = π₯
π.
This is defined for any π₯ β π .
β΄ domain of the function π¦ = πππ π₯ is the set of all reals, π .
We know that πππ π
2 = 0, πππ
3π
2 = 0.
By using πππ (2ππ + π) = πππ π we get
πππ (2ππ +π
2) = 0, β π β π.
This gives πππ 2π + 1 π
2 = 0, β π β π.
Thus πππ π₯ = 0 when π₯ = 2π + 1 π
2, π β π.
Since π ππ π₯ = 1
πππ π₯, the function π¦ = π ππ π₯ is defined only when
πππ π₯ β 0. Thus the domain of the function,
π¦ = π ππ π₯, is {π₯ βΆ π₯ β π , π₯ β 2π + 1 π
2, π β π }.
12. Find the domain of π(π₯) = π‘ππ π₯.
Solution: Recall the definition, π‘ππ π = π¦
π₯. This is defined for
any π₯, π¦ β π , π₯ β 0. We know that, π‘ππ π
2 = β, π‘ππ
3π
2 = β.
By using π‘ππ (2ππ + π) = π‘ππ π we get π‘ππ (2ππ +π
2) = β,
β π β π. This gives π‘ππ 2π + 1 π
2 = β, β π β π.
Thus π‘ππ π₯ = β when π₯ = 2π + 1 π
2, π β π. Thus the domain
of the function, π¦ = π‘ππ π₯, is {π₯ βΆ π₯ β π , π₯ β 2π + 1 π
2, π β π }.
13. Find the domain of π(π₯) = πππ‘ π₯.
Solution: Recall the definition, πππ‘ π = π₯
π¦. This is defined for
any π₯, π¦ β π , π¦ β 0. We know that πππ‘ π = β, πππ‘ 3π = β.
By using πππ‘ (2ππ + π) = πππ‘ π we get πππ‘ (2ππ + π) = β, β π β
π. This gives πππ‘ ππ = β, β π β π. Thus πππ‘ π₯ = β when
π₯ = ππ, π β π. Thus the domain of the function, π¦ = πππ‘ π₯, is
{π₯ βΆ π₯ β π , π₯ β ππ, π β π }.
14. Prove that π ππ (π₯ + π¦) = π ππ π₯. πππ π¦ + πππ π₯. π ππ π¦. Hence prove
that π ππ (π₯ β π¦) = π ππ π₯. πππ π¦ β πππ π₯. π ππ π¦.
Solution: π ππ (π₯ + π¦) = πππ π
2 β (π₯ + π¦)
= πππ π
2 β π₯ β π¦
= πππ π
2 β π₯ . πππ π¦ + π ππ
π
2 β π₯ . π ππ π¦
= π ππ π₯. πππ π¦ + πππ π₯. π ππ π¦.
By replacing π¦ by β π¦ we get,
π ππ (π₯ β π¦) = π ππ π₯. πππ (β π¦) + πππ π₯. π ππ ( β π¦)
= π ππ π₯. πππ π¦ + πππ π₯. (β π ππ π¦)
= π ππ π₯. πππ π¦ β πππ π₯. π ππ π¦.
15. Draw the graph of π¦ = π ππ π₯.
Solution: π ππ 0 = 0 and π ππ π 2 = 1.
β΄ π ππ π₯ increases from 0 to 1 as π₯ increases from 0 to π 2 .
π ππ π 2 = 1 and π ππ π = 0.
β΄ π ππ π₯ decreases from 1 to 0 as π₯ increases from π 2 to π.
π ππ π = 0 and π ππ 3π 2 = β 1.
β΄ π ππ π₯ decreases from 0 to β 1 as π₯ increases from π to 3π 2 .
π ππ 3π 2 = β1 and π ππ 2π = 0.
β΄ π ππ π₯ increases from β 1 to 0 as π₯ increases from 3π 2 to 2π.
Since π ππ π₯ is periodic with the period 2π we can make the
same observation in the
intervals
..., β4π,β 2π , β2π, 0 ,
0, 2π , 2π, 4π , 4π, 6π ,
...The graph is shown in the adjacent figure.
16. Draw the graph of π¦ = πππ π₯.
Solution: πππ 0 = 1 and πππ π 2 = 0.
β΄ πππ π₯ decreases from 1 to 0 as π₯ increases from 0 to π 2 .
πππ π 2 = 0 and πππ π = β 1.
β΄ πππ π₯ decreases from 0 to β 1 as π₯ increases from π 2 to π.
πππ π = β 1 and πππ 3π 2 = 0.
β΄ πππ π₯ increases from β 1 to 0 as π₯ increases from π to 3π 2 .
πππ 3π 2 = 0 and πππ 2π = 1.
β΄ πππ π₯ increases from 0 to 1 as π₯ increases from 3π 2 to 2π.
Since πππ π₯ is periodic with the period 2π, we can make the
same observation in the
intervals
..., β4π,β 2π , β2π, 0 ,
0, 2π , 2π, 4π , 4π, 6π ,
... The graph is shown in the adjacent figure.
17. Draw the graph of π¦ = π‘ππ π₯.
Solution: π‘ππ 0 = 0 and π‘ππ π 2 = β.
β΄ π‘ππ π₯ increases from 0 to β as π₯ increases from 0 to π 2 .
π‘ππ π 2 = β and π‘ππ π = 0.
Between π₯ = π 2 and π₯ = π, π‘ππ π₯ is negative.
β΄ π‘ππ π₯ increases from β β to 0 as π₯ increases from π 2 to
π.
Since π‘ππ π₯ is periodic with
the period π, we can make
the same observations in
the intervals, ... β2π, β π ,
βπ, 0 , 0, π , π, 2π , ...
The graph is shown in the
adjacent figure.
18. Draw the graph of π¦ = πππ‘ π₯.
Solution: πππ‘ 0 = β and πππ‘ π 2 = 0.
β΄ πππ‘ π₯ decreases from β to 0 as π₯ increases from 0 to π 2 .
πππ‘ π 2 = 0 and πππ‘ π = β.
Between π₯ = π 2 and π₯ = π πππ‘ π₯ is negative.
β΄ πππ‘ π₯ decreases from 0 to
β β as π₯ increases from π 2
to π.
Since πππ‘ π₯ is periodic with the
period π, we can make the
same observations in the
intervals, ... β2π, β π , βπ, 0 ,
0, π , π, 2π , ... The graph is shown in the adjacent figure.
19. Draw the graph of π¦ = πππ ππ π₯.
Solution: πππ ππ 0 = β and πππ ππ π 2 = 1.
β΄ πππ ππ π₯ decreases from β to 0 as π₯ increases from 0 to π 2 .
πππ ππ π 2 = 1 and πππ ππ π = β.
β΄ πππ ππ π₯ increases from 1 to β as π₯ increases from π 2 to π.
πππ ππ π = β and πππ ππ 3π 2 = β 1.
Between π₯ = π and π₯ = 3π 2 , πππ ππ π₯ is negative.
β΄ πππ ππ π₯ increases from
β β to β 1 as π₯ increases
from π to 3 π 2 .
πππ ππ 3π 2 = β 1 and
πππ ππ 2π = β.
Between π₯ = 3π 2 and
π₯ = 2π, πππ ππ π₯ is negative.
β΄ πππ ππ π₯ decreases from β 1 to β β as π₯ increases from
3 π 2 to 2π.
Since πππ ππ π₯ is periodic with the period 2π, we can make the
same observations in the intervals, ..., β4π, β 2π , β 2π, 0 ,
0, 2π , 2π, 4π , ... The graph is shown in the adjacent figure.
20. Draw the graph of π¦ = π ππ π₯.
Solution: π ππ 0 = 1 and π ππ π 2 = β.
β΄ π ππ π₯ increases from 1 to β as π₯ increases from 0 to π 2 .
π ππ π 2 = β and π ππ π = β 1.
Between π₯ = π 2 and π₯ = π, π ππ π₯ is negative.
β΄ π ππ π₯ increases from β β to β 1 as π₯ increases from π 2 to π.
π ππ π = β 1 and π ππ 3π 2 = β.
Between π₯ = π and π₯ = 3π 2 , π ππ π₯ is negative. β΄ π ππ π₯
decreases from β 1 to β β. as π₯ increases from π to 3 π 2 .
π ππ 3π 2 = β and π ππ 2π = 1.
Between π₯ = 3π 2 and
π₯ = 2π, π ππ π₯ is positive.
β΄ π ππ π₯ decreases from β
to 1 as π₯ increases from
3 π 2 to 2π.
Since π ππ π₯ is periodic with
the period 2π, we can make
the same observations in the
intervals, ..., β4π, β 2π , β 2π, 0 , 0, 2π , 2π, 4π , ... The
graph is shown in the adjacent figure.
21. If πππ π =4
5 and π is acute, find the value of
3πππ π + 2πππ ππ π
4 π ππ π β πππ‘ π .
Solution: Given cos π =4
5. Let πππ = 4 and βπ¦π = 5. Then
πππ = 3. Then πππ ππ π = βπ¦π
πππ =
5
3 and πππ‘ π =
πππ
πππ =
4
3.
Therefore 3πππ π + 2πππ ππ π
4 π ππ π β πππ‘ π =
3 4
5 + 2
5
3
4 3
5 β
4
3
= 12
5 +
10
312
5 β
4
3
= 86
1516
15
=86
16 =
43
8.
22. If π ππ π =13
5 and
3π
2< π < 2π, find the value of
2π ππ π β 3πππ π
4 π ππ π + 9 πππ π .
Solution: Given π ππ π =13
5. Let βπ¦π = 13 and πππ = 5.
Then πππ = 12. In the fourth quadrant π ππ π = β πππ
βπ¦π = β
12
13 ;
Thus 2π ππ π β 3πππ π
4 π ππ π + 9 πππ π =
2sin π β 3cos π
4 sin π 9 cos π=
2 β12
13 β 3
5
13
4 β12
13 + 9
5
13
=β24
13 β
15
13β48
13 +
45
13
=β39
β3= 13.
23. If π΄, π΅ are acute angles, π ππ π΄ =3
5 ; πππ π΅ =
12
13, find πππ π΄ + π΅
Solution: We have πππ π΄ + π΅ = πππ π΄ πππ π΅ β π πππ΄ π πππ΅.
Given π ππ π΄ =3
5 ; πππ π΅ =
12
13.
β΄ πππ π΄ = 1 β π ππ2π΄ = 1 β9
25 =
25 β 9
25 =
4
5 and
π πππ΅ = 1 β πππ 2π΅ = 1 β144
169 =
5
13.
β΄ πππ π΄ + π΅ = 4
5Γ
12
13 β
3
5Γ
5
13 =
48 β 15
65 =
33
65.
24. If π΄ + π΅ = 45Β° then prove that 1 + π‘ππ π΄ 1 + π‘ππ π΅ = 2 and
hence deduce that π‘ππ 221
2
Β°= 2 β 1.
Solution: Given π΄ + π΅ = 45Β° βΉ π‘ππ π΄ + π΅ = π‘ππ 45Β°
βΉ π‘ππ π΄ + π‘ππ π΅
1 β π‘ππ π΄ π‘ππ π΅ = 1 βΉ π‘ππ π΄ + π‘ππ π΅ = 1 β π‘ππ π΄. π‘ππ π΅
βΉ π‘ππ π΄ + π‘ππ π΅ + π‘ππ π΄. π‘ππ π΅ = 1 (by adding 1 to both sides)
βΉ 1 + π‘ππ π΄ + π‘ππ π΅ + π‘ππ π΄. π‘ππ π΅ = 2
βΉ 1 + π‘ππ π΄ + π‘ππ π΅. 1 + π‘ππ π΄ = 2
βΉ 1 + π‘ππ π΄ . 1 + π‘ππ π΅ = 2.
By taking π΄ = π΅ = 221
2
Β° we get,
1 + π‘ππ 221
2
Β°
2
= 2 βΉ 1 + π‘ππ 221
2
Β°= Β± 2
βΉ π‘ππ 221
2
Β°= Β± 2 β 1.
Since 221
2
Β° is acute, π‘ππ 221
2
Β° is positive. Therefore
π‘ππ 221
2
Β°= 2 β 1.
25. Prove that π ππ 3π₯ = 3π ππ π₯ β 4π ππ3 π₯
Solution: π ππ 3π₯ = π ππ (2π₯ + π₯) = π ππ 2π₯. πππ π₯ + πππ 2π₯. π ππ π₯
= (2π ππ π₯. πππ π₯)πππ π₯ + (1 β 2π ππ2 π₯)π ππ π₯
= 2. π ππ π₯. πππ 2 π₯ + π ππ π₯ β 2π ππ3 π₯
= 2. π ππ π₯(1 β π ππ2 π₯) + π ππ π₯ β 2π ππ3 π₯ = 3π ππ π₯ β 4π ππ3 π₯.
26. Prove that πππ 3π₯ = 4πππ 3 π₯ β 3πππ π₯
Solution: πππ 3π₯ = πππ (2π₯ + π₯) = πππ 2π₯. πππ π₯ β π ππ 2π₯. π ππ π₯
= (2πππ 2π₯ β 1)πππ π₯ β (2π ππ π₯. πππ π₯)π ππ π₯
= 2πππ 3 π₯ β πππ π₯ β 2π ππ2 π₯. πππ π₯
= 2πππ 3 π₯ β πππ π₯ β 2(1 β πππ 2 π₯)πππ π₯
= 2πππ 3 π₯ β πππ π₯ β 2πππ π₯ + 2πππ 3 π₯ = 4πππ 3 π₯ β 3πππ π₯.
27. Prove that π‘ππ 3π₯ = 3π‘ππ π₯ β π‘ππ 3 π₯
1 β 3π‘ππ 2 π₯
Solution: π‘ππ 3π₯ = π‘ππ (2π₯ + π₯) =π‘ππ 2π₯ + π‘ππ π₯
1 β π‘ππ 2π₯ .π‘ππ π₯
= 2π‘ππ π₯
1 β π‘ππ 2 π₯ + π‘ππ π₯
1 β 2π‘ππ π₯
1 β π‘ππ 2 π₯ .π‘ππ π₯
= 2π‘ππ π₯ + π‘ππ π₯ (1 β π‘ππ 2 π₯)
1 β π‘ππ 2 π₯ β 2π‘ππ 2 π₯
= 2π‘ππ π₯ + π‘ππ π₯ β π‘ππ 3 π₯)
1 β π‘ππ 2 π₯ β 2π‘ππ 2 π₯ =
3π‘ππ π₯ β π‘ππ 3 π₯
1 β 3π‘ππ 2 π₯.
28. If π‘ππ π΄ =1 β πππ π΅
π ππ π΅, prove that π‘ππ 2π΄ = π‘ππ π΅, where π΄ and π΅
are acute angles.
Solution: π π»π =1 β πππ π΅
π ππ π΅ =
2π ππ 2 π΅
2
2 π ππ π΅2 βπππ
π΅
2
=π ππ π΅
2
πππ π΅
2
= π‘ππ π΅2 .
β΄ π‘ππ π΄ = π‘ππ π΅2 .
Since π΄ and π΅ are acute angles we get, π΄ = π΅
2 . β΄ 2π΄ = π΅
β΄ π‘ππ 2π΄ = π‘ππ π΅.
29. Show that 4 π ππ π΄. π ππ 60Β° + π΄ . π ππ 60Β° β π΄ = π ππ 3π΄.
Solution: πΏπ»π = 4 π ππ π΄. {π ππ 60Β° + π΄ . π ππ 60Β° β π΄ }
= 4 π ππ π΄ {π ππ2 60Β° β π ππ2 π΄} = 4 π ππ π΄ 3
2
2
β π ππ2 π΄
= 4 π ππ π΄ 3
4β π ππ2π΄ = 3π ππ π΄ β 4π ππ3π΄ = π ππ3π΄ = π π»π.
30. Show that 4 πππ π΄. πππ 60Β° + π΄ . πππ 60Β° β π΄ = πππ 3π΄.
Solution: πΏπ»π = 4 πππ π΄. πππ 60Β° + π΄ . πππ 60Β° β π΄
= 4 πππ π΄ {πππ 2 60Β° β π ππ2 π΄}
= 4 πππ π΄ 1
4β 1 β πππ 2π΄ = 4 πππ π΄ β
3
4+ πππ 2π΄
= β 3πππ π΄ + 4πππ 3π΄ = 4πππ 3π΄ β 3πππ π΄ = πππ 3π΄ = π π»π.
31. Prove that 2 π ππ π΄ β π ππ 2π΄
2 π ππ π΄ + π ππ 2π΄ = π‘ππ2
π΄
2.
Solution: 2 π ππ π΄ β π ππ 2π΄
2 π ππ π΄ + π ππ 2π΄ =
2 π ππ π΄ β 2π ππ π΄.πππ π΄
2 π ππ π΄ + 2π ππ π΄.πππ π΄
= 2 π ππ π΄(1 β πππ π΄)
2 π ππ π΄(1 + πππ π΄) =
2π ππ 2 π΄
2
2πππ 2 π΄
2
= π‘ππ2 π΄
2.
32. Prove that π ππ 4π₯ πππ 2π₯
1 + πππ 4π₯ 1 + πππ 2π₯ = π‘ππ π₯
Solution: π ππ 4π₯ πππ 2π₯
1 + πππ 4π₯ 1 + πππ 2π₯ =
2π ππ 2π₯ .πππ 2π₯ .πππ 2π₯
2πππ 2 2π₯ .2πππ 2 π₯ =
π ππ 2π₯
2πππ 2 π₯
= 2π ππ π₯ .πππ π₯
2πππ π₯ .πππ π₯ = π‘ππ π₯.
33. If π ππ π΄ = 3
5 and π΄ is acute angle, find the values of π ππ 2π΄
and πππ 2π΄.
Solution: π ππ π΄ = 3
5 β πππ = 3 and βπ¦π = 5. β΄ πππ = 4.
β΄ πππ π΄ = πππ
βπ¦π=
4
5.
β΄ π ππ 2π΄ = 2π ππ π΄. πππ π΄ = 23
5.
4
5 =
24
25 ;
πππ 2π΄ = 1 β 2π ππ2π΄ = 1 β 2 3
5
2 = 1 β 2
9
25 = 1 β
18
25 =
7
25.
34. If π‘ππ π΄ = 1
3, π‘ππ π΅ =
1
7 prove that 2π΄ + π΅ =
π
4.
Solution: π‘ππ 2π΄ = 2π‘ππ π΄
1 β π‘ππ 2 π΄ =
2 1 3
1 β 1 3 2 =
2 3
1 β 1 9 =
2 3
8 9 =
3
4.
β΄ π‘ππ (2π΄ + π΅) = π‘ππ 2π΄ + π‘ππ π΅
1 β π‘ππ 2π΄.π‘ππ π΅ =
3
4 +
1
7
1 β 3
4
1
7
= 21 + 4
2828 β 3
28
= 25
25 = 1.
Since π‘ππ π΄ and π‘ππ π΅ are positive and since π‘ππ (2π΄ + π΅) is
positive, we get, 2π΄ + π΅ = π
4.
35. Show that π‘ππ π΄ =1 β πππ 2π΄
π ππ 2π΄ and hence deduce the value of
π‘ππ15Β°.
Solution: 1 β πππ 2π΄
π ππ 2π΄ =
2π ππ 2 π΄
2π ππ π΄.πππ π΄ = π‘ππ π΄. β΄ π‘ππ π΄ =
1 β πππ 2π΄
π ππ 2π΄.
By taking π΄ = 15Β° we get,
π‘ππ 15Β° =1 β πππ 2(15Β°)
π ππ 2(15Β°) =
1 β πππ 30Β°
π ππ 30Β° =
1 β 3 2
1 2 =
2 β 3
2
1 2 = 2 β 3.
36. Prove that πππ 20Β° + πππ 100Β° + πππ 140Β° = 0.
Solution: πΏπ»π = πππ 20Β° + πππ 100Β° + πππ 140Β°
= πππ 20Β° + 2 πππ 100 Β° + 140 Β°
2 . πππ
100Β° β 140Β°
2
= πππ 20Β° + 2 πππ 120Β° . πππ β20Β°
= πππ 20Β° + 2 β1
2 . πππ 20Β° = πππ 20Β° β πππ 20Β° = 0.
37. Prove that π ππ 50Β° β π ππ 70Β° + π ππ 10Β° = 0
Solution: πΏπ»π = π ππ 50Β° β π ππ 70Β° + π ππ 10Β°
= 2 πππ 50 Β° + 70Β°
2 . π ππ
50Β° β 70Β°
2 + π ππ10Β°
= 2 πππ 60Β° β π ππ β10Β° + π ππ10Β°
= β 2 Γ1
2π ππ 10Β° + π ππ 10Β° = β π ππ 10Β° + π ππ 10Β° = 0.
38. Prove that π ππ 20Β° π ππ 40Β° π ππ 60Β° π ππ 80Β° =3
16
Solution: πΏπ»π = π ππ 60Β° . π ππ 20Β° π ππ(60Β° β 20Β°) π ππ(60Β° + 20Β°)
= 3
2π ππ 20Β° π ππ 60Β° β 20Β° π ππ 60Β° + 20Β°
= 3
2.
1
4. sin (3 Γ 20Β°) =
3
2.
1
4. sin 60Β°
= 3
8 π ππ 60Β° =
3
8Γ
3
2=
3
16.
39. Show that π ππ 5π΄ + π ππ 2π΄ β π ππ π΄
πππ 5π΄ + πππ 2π΄ + πππ π΄ = π‘ππ 2π΄
Solution: π ππ 5π΄ + π ππ 2π΄ β π ππ π΄
πππ 5π΄ + πππ 2π΄ + πππ π΄=
π ππ 2π΄ + π ππ 5π΄ β π ππ π΄
πππ 2π΄ + πππ 5π΄ + πππ π΄
=π ππ 2π΄ + 2 πππ 5π΄ + π΄
2 .π ππ
5π΄ β π΄
2
πππ 2π΄ + 2 πππ 5π΄ + π΄2
.πππ 5π΄ β π΄
2
=π ππ 2π΄ + 2 πππ 3π΄ . π ππ 2π΄
πππ 2π΄ + 2 πππ 3π΄ . πππ 2π΄
=π ππ 2π΄[1 + 2 πππ 3π΄]
πππ 2π΄[1 + 2 πππ 3π΄] =
π ππ 2π΄
πππ 2π΄= π‘ππ 2π΄.
40. Find the general solution of π ππ π = π, β1 β€ π β€ 1.
Solution: Let πΌ be an angle such that
π ππ πΌ = π, βπ
2 β€ πΌ β€
π
2.
Then π ππ π = π ππ πΌ.
Now π ππ π = π ππ πΌ β π ππ π β π ππ πΌ = 0
β 2πππ π + πΌ
2. π ππ
π β πΌ
2 = 0
β πππ π + πΌ
2 = 0 or π ππ
π β πΌ
2 = 0.
Now πππ π + πΌ
2 = 0 β
π + πΌ
2 = (2π + 1)
π
2, π β π.
β π + πΌ = (2π + 1)π
β π = (2π + 1)π β πΌ β¦(1)
π ππ π β πΌ
2 = 0 β
π β πΌ
2 = ππ, π β π.
β π β πΌ = 2ππ β π = 2ππ + πΌ β¦(2).
β΄ combining (1) and (2),we get, π = ππ + β 1 ππΌ, π β π.
This is the general solution of π ππ π = π, β 1 β€ π β€ 1.
41. Find the general solution of πππ π = π, β 1 β€ π β€ 1.
Solution: Let πΌ be an angle such that πππ πΌ = π, 0 β€ πΌ β€ π
Then πππ π = πππ πΌ.
Now πππ π = πππ πΌ β πππ π β πππ πΌ = 0
β β 2π ππ π + πΌ
2. π ππ
π β πΌ
2 = 0
β π πππ + πΌ
2 = 0 or π ππ
π β πΌ
2 = 0.
Now π ππ π + πΌ
2 = 0 β
π + πΌ
2 = ππ, π β π
β π + πΌ = 2ππ β π = 2ππ β πΌ β¦(1)
π ππ π β πΌ
2 = 0 β
π β πΌ
2 = ππ, π β π
β π β πΌ = 2ππ β π = 2ππ + πΌ ...(2).
β΄ combining (1) and (2), we get, π = 2ππ Β± πΌ , π β π.
This is the general solution of πππ π = π, β 1 β€ π β€ 1.
42. Find the general solution of π‘ππ π = π, β β < π < β.
Solution: Let πΌ be an angle such that
π‘ππ πΌ = π , β π
2 < πΌ <
π
2.
Then π‘ππ π = π‘ππ πΌ.
Now π‘ππ π = π‘ππ πΌ β π πππ
πππ π =
π πππΌ
πππ πΌ
β π ππ π. πππ πΌ = πππ π. π ππ πΌ
β π ππ π. πππ πΌ β πππ π. π ππ πΌ = 0
β π ππ( π β πΌ ) = 0 β π β πΌ = ππ, π β π.
β π = ππ + πΌ , π β π.
This is the general solution of π‘ππ π = π, β β < π < β.
43. Find the general solution of π‘ππ 2π. π‘ππ 3π = 1.
Solution: π‘ππ 2π. π‘ππ 3π = 1 βΉ π ππ 2 π
πππ 2 π.π ππ 3 π
πππ 3 π= 1
βΉ π ππ 2π. π ππ 3π = πππ 2π. πππ 3π
βΉ πππ 2π. πππ 3π β π ππ 2π. π ππ 3π = 0
βΉ πππ (2π + 3π) = 0 βΉ πππ 5π = 0.
βΉ 5 π = 2π + 1 π
2, π β π βΉ π = 2π + 1
π
10, π β π.
This is the general solution of the given equation.
44. Find the general solution of
π‘ππ π + π‘ππ 2π + 3π‘ππ π. π‘ππ 2π = 3.
Solution: π‘ππ π + π‘ππ 2π + 3π‘ππ π. π‘ππ 2π = 3
βΉ π‘ππ π + π‘ππ 2π = 3(1 β π‘ππ π. π‘ππ 2π)
βΉ π‘ππ π + π‘ππ 2π
(1 β π‘ππ πβπ‘ππ 2π) = 3 βΉ π‘ππ 3π = 3 = π‘ππ
π
3
βΉ 3π = ππ + π
3 , π β π βΉ π =
ππ
3 +
π
9 , π β π.
This is the general solution of the given equation.
45. Find the general solution of π ππ 2π₯ + π ππ 4π₯ + π ππ 6π₯ = 0.
Solution: π ππ 2π₯ + π ππ 4π₯ + π ππ 6π₯ = 0
βΉ π ππ 4π₯ + π ππ 6π₯ + π ππ 2π₯ = 0
βΉ π ππ 4π₯ + 2π ππ 6π₯ + 2π₯
2 β πππ
6π₯ β 2π₯
2 = 0
βΉ π ππ 4π₯ + 2π ππ 4π₯. πππ 2π₯ = 0 βΉ π ππ 4π₯(1 + 2πππ 2π₯) = 0
βΉ π ππ 4π₯ = 0 or 1 + 2πππ 2π₯ = 0.
π ππ 4π₯ = 0 βΉ 4π₯ = ππ β΄ π₯ = ππ
4, π β π.
1 + 2πππ 2π₯ = 0 βΉ πππ 2π₯ = β1
2 = πππ
2π
3.
β΄ 2π₯ = 2ππ Β± 2π
3, π β π. β΄ π₯ = ππ Β±
π
3, π β π.
This is the general solution of the given equation.
46. Find the general solution of 4π ππ2 π₯ β 8πππ π₯ + 1 = 0.
Solution: 4π ππ2 π₯ β 8πππ π₯ + 1 = 0
βΉ 4(1 β πππ 2 π₯) β 8πππ π₯ + 1 = 0
βΉ β4πππ 2 π₯ β 8πππ π₯ + 5 = 0
βΉ 4πππ 2 π₯ + 8πππ π₯ β 5 = 0
βΉ 4πππ 2 π₯ + 10πππ π₯ β 2πππ π₯ β 5 = 0
βΉ 2πππ π₯ 2πππ π₯ + 5 β 1 2πππ π₯ + 5 = 0
βΉ 2πππ π₯ β 1 2πππ π₯ + 5 = 0
βΉ 2πππ π₯ β 1 = 0 or 2πππ π₯ + 5 = 0
βΉ πππ π₯ =1
2 or πππ π₯ = β
5
2
Now πππ π₯ = β 5
2< β1. This has no solution.
πππ π₯ =1
2 βΉ π₯ = 2ππ Β±
π
3 , π β π.
This is the general solution of the given equation.
47. Find the general solution of 2π ππ2 π₯ + 3πππ π₯ + 1 = 0.
Solution: 2π ππ2 π₯ + 3πππ π₯ + 1 = 0
βΉ 2(1 β πππ 2 π₯) + 3πππ π₯ + 1 = 0
βΉ 2πππ 2 π₯ β 2 3πππ π₯ + 3πππ π₯ β 3 = 0
βΉ 2πππ π₯ πππ π₯ β 3 + 3(πππ π₯ β 3) = 0
βΉ (2πππ π₯ + 3) πππ π₯ β 3 = 0
βΉ πππ π₯ = β 3
2 or πππ π₯ = 3.
πππ π₯ = 3 > 1. This has no solution.
Now πππ π₯ = β 3
2 = cos π β
π
6 = πππ
5π
6. β΄ π₯ = 2ππ Β±
5π
6, π β π.
This is the general solution of the given equation.
48. Find the general solution of 2πππ 2 π₯ + 3π ππ π₯ = 0.
Solution: 2πππ 2 π₯ + 3π ππ π₯ = 0
βΉ 2 1 β π ππ2 π₯ + 3π ππ π₯ = 0 βΉ 2 β 2π ππ2 π₯ + 3π ππ π₯ = 0
βΉ 2π ππ2 π₯ β 3 π ππ π₯ β 2 = 0 βΉ 2π ππ π₯ + 1 π ππ π₯ β 2 = 0
βΉ 2π ππ π₯ + 1 = 0 or π ππ π₯ β 2 = 0
βΉ π ππ π₯ = β 1
2 or π ππ π₯ = 2
π πππ₯ = 2 > 1 has no solution.
π ππ π₯ = β 1
2 = βπ ππ
π
6= π ππ β
π
6 .
β΄ π₯ = ππ + (β1)π βπ
6 , π β π.
This is the general solution of the given equation.
49. Find the general solution of 2πππ 2 π + π ππ π = π ππ 2π + πππ π.
Solution: 2πππ 2 π + π ππ π = π ππ 2π + πππ π
βΉ 2πππ 2 π β πππ π = 2π ππ ππππ π β π ππ π
βΉ πππ π 2πππ π β 1 β π ππ π 2πππ π β 1 = 0
βΉ 2πππ π β 1 (πππ π β π ππ π) = 0
βΉ 2πππ π β 1 = 0 or πππ π β π ππ π = 0.
Now 2πππ π β 1 = 0 βΉ πππ π = 1
2 = πππ
π
3 β΄ π = 2ππ Β±
π
3 , π β π.
πππ π β π ππ π = 0 βΉ π ππ π = πππ π.
Dividing by πππ π we get, π‘ππ π = 1 = π‘πππ
4. β΄ π = ππ +
π
4,
π β π. β΄ the general solution is π = 2ππ Β± π
3, π = ππ +
π
4, π β π.
50. Find the general solution of πππ π₯ + πππ 2π₯ + πππ 3π₯ = 0.
Solution: πππ π₯ + πππ 2π₯ + πππ 3π₯ = 0
βΉ 2 πππ π₯ + 3π₯
2 . πππ
π₯ β 3π₯
2 + πππ 2π₯ = 0 βΉ πππ 2π₯ 2πππ π₯ + 1 = 0
βΉ πππ 2π₯ = 0 or 2πππ π₯ + 1 = 0.
Now πππ 2π₯ = 0 βΉ 2π₯ = (2π + 1)π
2 β΄ π₯ = 2π + 1
π
4 ,π β π.
2πππ π₯ + 1 = 0 βΉ πππ π₯ = β1
2= πππ π β
π
3 = πππ
2π
3
β΄ π₯ = 2ππ Β±2π
3 , π β π.
β΄ the general solution is π₯ = 2π + 1 π
4, π₯ = 2ππ Β±
2π
3 π β π.
51. Prove that 2πππ π
13. πππ
9π
13 + πππ
3π
13 + πππ
5π
13 = 0.
Solution: 2πππ π
13. πππ
9π
13 + πππ
3π
13 + πππ
5π
13
= 2πππ π
13. πππ
9π
13 + 2πππ
3π
13 +
5π
13
2. πππ
5π
13 β
3π
13
13
= 2πππ π
13. πππ
9π
13 + 2πππ
8π
13
2. πππ
2π
13
2
= 2πππ π
13. πππ
9π
13 + 2πππ
4π
13. πππ
π
13
= 2πππ π
13. πππ
13π β 4π
13 + 2πππ
4π
13. πππ
π
13
= 2πππ π
13. πππ π β
4π
13 + 2πππ
4π
13. πππ
π
13
= β 2πππ π
13. πππ
4π
13 + 2πππ
4π
13. πππ
π
13 = 0.
52. Prove that πππ 6π₯ = 32πππ 6 π₯ β 48πππ 4 π₯ + 18πππ 2 π₯ β 1
Solution: πππ 6π₯ = 2πππ 2 3π₯ β 1
= 2 4πππ 3 π₯ β 3πππ π₯ 2 β 1
= 2 4πππ 3 π₯ 2 + 3πππ π₯ 2 β 24πππ 3 π₯. πππ π₯ β 1
= 2 16πππ 6 π₯ + 9πππ 2 π₯ β 24πππ 4 π₯ β 1
= 32πππ 6 π₯ β 48πππ 4 π₯ + 18πππ 2 π₯ β 1
53. Prove that πππ‘ π₯. πππ‘ 2π₯ β πππ‘ 2π₯. πππ‘ 3π₯ β πππ‘ 3π₯. πππ‘ π₯ = 1.
Solution: We have π₯ + 2π₯ = 3π₯.
β΄ πππ‘ 3π₯ = πππ‘ (π₯ + 2π₯) = πππ‘ π₯ .πππ‘ 2π₯ β 1
πππ‘ 2π₯ + πππ‘ π₯.
β΄ πππ‘ 3π₯(πππ‘ 2π₯ β πππ‘ π₯) = πππ‘ π₯. πππ‘ 2π₯ + 1
β΄ πππ‘ 3π₯. πππ‘ 2π₯ + πππ‘ 3π₯. πππ‘ π₯ = πππ‘ π₯. πππ‘ 2π₯ β 1
Thus πππ‘ 3π₯. πππ‘ 2π₯ + πππ‘ 3π₯. πππ‘ π₯ β πππ‘ π₯. πππ‘ 2π₯ = β 1.
By changing the sign we get
πππ‘ π₯. πππ‘ 2π₯ β πππ‘ 2π₯. πππ‘ 3π₯ β πππ‘ 3π₯. πππ‘ π₯ = 1.
Five mark questions
1. Discuss the signs of trigonometric functions in the four
quadrants.
Solution: Consider a circle of radius π,
centered at the origin of a Cartesian
coordinate system. Let π(π₯, π¦) be a
point on the circle. Suppose that the
angle made by the radius ππ with the
positive direction of the
x-axis is π. Then
π ππ π = π¦
π, πππ π =
π₯
π, π‘ππ π =
π¦
π₯ .
a) If 0 < π < π
2 then π(π₯, π¦ ) is a point in the first quadrant.
Here π₯ > 0 and π¦ > 0. β΄ π ππ π > 0, πππ π > 0, π‘ππ π > 0
β΄ πππ ππ π > 0, π ππ π > 0, πππ‘ π > 0.
In the first quadrant πππ the trigonometric functions are
positive.
b) If π
2 < π < π then π(π₯,π¦ ) is a point in the second quadrant.
Here π₯ < 0 and π¦ > 0. β΄ π ππ π > 0, πππ π < 0 , π‘ππ π < 0
β΄ πππ ππ π > 0, π ππ π < 0, πππ‘ π < 0.
In the second quadrant πππ π½ (and also ππππ π½ ) is positive.
c) If π < π < 3π
2 then π(π₯,π¦ ) is a point in the third quadrant.
Here π₯ < 0 and π¦ < 0. β΄ π ππ π < 0, πππ π < 0, π‘ππ π > 0
β΄ πππ ππ π < 0, π ππ π < 0, πππ‘ π > 0.
In the third quadrant πππ π½ ( and also πππ π½ ) is positive.
d) If 3π
2 < π < 2π then π(π₯,π¦) is a point in the
fourth quadrant. Here π₯ > 0 and π¦ < 0.
β΄ π ππ π < 0, πππ π > 0 and π‘ππ π < 0.
β΄ πππ ππ π < 0, π ππ π > 0 , πππ‘ π < 0.
In the fourth quadrant πππ π½ (and also ππ π½ ) is positive.
2. Find the range of values of the trigonometric functions
Solution: Consider a circle of radius π, centered at the origin
of a Cartesian coordinate system.
Let π(π₯, π¦) be a point on the circle.
Suppose that the angle made by the
radius ππ with the positive direction
of the x-axis is π.
Then π ππ π = π¦
π , πππ π =
π₯
π ,
π‘ππ π = π¦
π₯ ,
πππ ππ π = π
π¦ , π ππ π =
π
π₯ , πππ‘ π =
π₯
π¦.
For any position of the point π(π₯, π¦) on the circle,
β π β€ π¦ β€ π β β 1 β€ π¦/π β€ 1 β β 1 β€ π πππ β€ 1.
Similarly
β π β€ π₯ β€ π β β 1 β€ π₯/π β€ 1 β β 1 β€ πππ π β€ 1.
Since β β < π¦/π₯ < β and β β < π₯/π¦ < β, we get,
β β < π‘πππ < β and β β < πππ‘π < β.
Also, β π β€ π¦ β€ π β π/π¦ β₯ 1 or π/π¦ β€ β 1
β΄ | π/π¦ | β₯ 1. β΄ |π πππ | β₯ 1.
Similarly β π β€ π₯ β€ π β π/π₯ β₯ 1 or π/π₯ β€ β 1
β΄ | π/π₯ | β₯ 1. β΄ |πππ πππ | β₯ 1.
3. Express the trigonometric ratios of β π with those of π.
Solution: Consider a circle of radius π centered at the origin of
the Cartesian coordinate system. Let
π( π₯, π¦ ) be a point on the circle. Suppose
that the angle made by the radius ππ
with the positive direction of the x-axis is
π. Then
π ππ π = π¦
π , πππ π =
π₯
π , π‘ππ π =
π¦
π₯.
Draw ππ perpendicular to the x-axis and
produce to meet the circle at π.
Here π is the image of π. β΄ π β‘ ( π₯, β π¦ ). Here angle πππ is
β π.
By definition,
π ππ (β π) = β π¦
π = β
π¦
π = β π ππ π ;
πππ (β π) = π₯
π = πππ π ;
π‘ππ (β π) = β π¦
π₯ = β
π¦
π₯ = β π‘ππ π.
By writing the reciprocals we get πππ ππ (β π) = β πππ ππ π ;
π ππ (β π) = π ππ π ; πππ‘ (β π) = β πππ‘ π.
4. Prove geometrically πππ (π₯ + π¦) = πππ π₯. πππ π¦ β π ππ π₯. π ππ π¦. Hence
prove that πππ (π₯ β π¦) = πππ π₯. πππ π¦ + π ππ π₯. π ππ π¦
Solution: Consider the unit circle centered at the origin of the Cartesian
coordinate system.
Let π and π be points on the unit circle such that angle π = π₯ , angle
πππ = π¦ and angle πππ = π₯ + π¦.
Then π β‘ (πππ π₯, π ππ π₯) and π β‘ (πππ (π₯ + π¦), π ππ (π₯ + π¦)).
Let π be a point on the unit circle such that angle πππ = β π¦.
Then π β‘ (πππ (β π¦), π ππ (β π¦)). Let π β‘ (1, 0).
In βπππ and βπππ, ππ = ππ = ππ = ππ.
Also angle πππ = πππ.
β΄ triangles are congruent β ππ = ππ β ππ 2 = ππ2.
β πππ π₯ β πππ (β π¦) 2 + π ππ π₯ β π ππ(β π¦) 2
= πππ (π₯ + π¦) β 1 2 + π ππ (π₯ + π¦) β 0 2
β πππ π₯ β πππ π¦ 2 + π ππ π₯ + π ππ π¦ 2
= πππ (π₯ + π¦) β 1 2 + π ππ (π₯ + π¦) 2
β πππ 2 π₯ + πππ 2 π¦ β 2. πππ π₯. πππ π¦ + π ππ2 π₯ + π ππ2 π¦ + 2. π ππ π₯. π ππ π¦
= πππ 2 (π₯ + π¦) + 1 β 2. πππ (π₯ + π¦) + π ππ2 (π₯ + π¦)
β 2 β 2. πππ π₯. πππ π¦ + 2. π ππ π₯. π ππ π¦ = 2 β 2. πππ (π₯ + π¦)
β β πππ π₯. πππ π¦ + π ππ π₯. π ππ π¦ = β πππ (π₯ + π¦)
β πππ (π₯ + π¦) = πππ π₯. πππ π¦ β π ππ π₯. π ππ π¦.
By replacing π¦ by β π¦ we get
πππ (π₯ + (βπ¦)) = πππ π₯. πππ (βπ¦) β π ππ π₯. π ππ (βπ¦)
β πππ (π₯ β π¦) = πππ π₯. πππ π¦ β π ππ π₯. (β π ππ π¦)
= πππ π₯. πππ π¦ + π ππ π₯. π ππ π¦.
5. Express the trigonometric ratios of 90Β° β π with those of π
Solution: We know that πππ (π₯ + π¦) = πππ π₯. πππ π¦ β π ππ π₯. π ππ π¦.
a) By taking π₯ = 90Β° and π¦ = β π we get,
πππ (90Β° β π) = πππ 90Β°. πππ (β π) β π ππ 90Β°. π ππ ( βπ)
= (0). πππ π β (1)(βπ ππ π) = π ππ π.
β΄ πππ (90Β° β π) = π ππ π.
b) Replacing π by 90Β° β π in π ππ π = πππ (90Β° β π) we get,
π ππ (90Β° β π) = πππ 90Β° β (90Β° β π) = πππ π.
c) π‘ππ (90Β° β π) = π ππ (90Β° βπ)
πππ (90Β° β π) =
πππ π
π ππ π = πππ‘ π.
d) πππ‘ (90Β° β π) = πππ (90Β° βπ)
π ππ (90Β° β π) =
π ππ π
πππ π = π‘ππ π
e) πππ ππ (90Β° β π) = 1
π ππ (90Β° βπ) =
1
πππ π = π ππ π.
f) π ππ (90Β° β π) = 1
πππ (90Β° βπ) =
1
π ππ π = πππ ππ π.
6. Express the trigonometric ratios of 90Β° + π with those of π.
Solution: We know that πππ (π₯ + π¦) = πππ π₯. πππ π¦ β π ππ π₯. π ππ π¦.
By taking π₯ = 90Β° and π¦ = π we get,
a) πππ (90Β° + π) = πππ 90Β°. πππ π β π ππ 90Β°. π ππ π
= (0). πππ π β (1). π ππ π = β π ππ π.
β΄ πππ (90Β° + π) = β π ππ π.
b) π ππ (90Β° + π) = πππ 90Β° β (90Β° + π) = πππ (β π) = πππ π.
c) π‘ππ (90Β° + π) = π ππ (90Β° + π)
πππ (90Β° + π) =
πππ π
β π ππ π = β πππ‘ π.
d) πππ‘ (90Β° + π) = πππ (90Β° + π)
π ππ (90Β° + π) =
β π ππ π
πππ π = β π‘ππ π
e) πππ ππ (90Β° + π) = 1
π ππ (90Β° + π) =
1
πππ π = π ππ π.
f) π ππ (90Β° + π) = 1
πππ (90Β° + π) =
1
β π ππ π = β πππ ππ π.
7. Find the value of π ππ 18Β°.
Hence find πππ 36Β°.
Solution: Let π = 18Β°. Then 5π = 90Β°.
β΄ 3π + 2π = 90Β° or 2π = 90Β° β 3π.
β΄ π ππ 2π = π ππ (90Β° β 3π) = πππ 3π.
β΄ 2π ππ π. πππ π = 4πππ 3 π β 3πππ π.
Dividing by πππ π, we get, 2π ππ π = 4πππ 2 π β 3
β 2π ππ π = 4(1 β π ππ2 π) β 3
β 2π ππ π = 4 β 4π ππ2 π β 3
β 2π ππ π = 1 β 4π ππ2 π
β 4π ππ2 π + 2π ππ π β 1 = 0
β π ππ π = β 2 Β± 4 β 4(4)(β 1)
2 Γ 4 =
β 2 Β± 4 + 16
8 =
β 2 Β± 20
8
= β 2 Β± 2 5
8 =
2(β 1 Β± 5)
8 =
β 1 Β± 5
4.
Since π = 18Β°, π ππ π = π ππ 18Β° is positive. β΄ π ππ 18Β° = 5 β 1
4.
By using πππ 2π΄ = 1 β 2π ππ2π΄, we get,
πππ 36Β° = 1 β 2π ππ2 18Β° = 1 β 2 5 β 1
4
2
= 1 β 2 5 + 1 β 2 5
16 = 1 β
6 β 2 5
8 =
8 β (6 β 2 5)
8
= 2 + 2 5
8 =
1 + 5
4.
β΄ πππ 36Β° = 5 + 1
4.
8. Find the value of π ππ 18Β°.
Hence find πππ 18Β°
Solution: Let π = 18Β°. Then 5π = 90Β°.
β΄ 3π + 2π = 90Β° or 2π = 90Β° β 3π.
β΄ π ππ 2π = π ππ (90Β° β 3π) = πππ 3π.
β΄ 2π ππ π. πππ π = 4πππ 3 π β 3πππ π.
Dividing by πππ π, we get, 2π ππ π = 4πππ 2 π β 3
β 2π ππ π = 4(1 β π ππ2 π) β 3
β 2π ππ π = 4 β 4π ππ2 π β 3
β 2π ππ π = 1 β 4π ππ2 π
β 4π ππ2 π + 2π ππ π β 1 = 0
β π ππ π = β 2 Β± 4 β 4(4)(β 1)
2 Γ 4 =
β 2 Β± 4 + 16
8 =
β 2 Β± 20
8
= β 2 Β± 2 5
8 =
2(β 1 Β± 5)
8 =
β 1 Β± 5
4.
Since π = 18Β°, π ππ π = π ππ 18Β° is positive. β΄ π ππ 18Β° = 5 β 1
4.
By using πππ 2π΄ = 1 β 2π ππ2π΄, we get,
πππ 36Β° = 1 β 2π ππ2 18Β° = 1 β 2 5 β 1
4
2
= 1 β 2 5 + 1 β 2 5
16 = 1 β
6 β 2 5
8 =
8 β (6 β 2 5)
8
= 2 + 2 5
8 =
1 + 5
4.
β΄ πππ 36Β° = 5 + 1
4.
By using 2πππ 2π΄ = 1 + πππ 2π΄, we get,
2πππ 218Β° = 1 + πππ 36Β° = 1 + 5 + 1
4 =
4 + 5 + 1
4 =
5 + 5
4
β΄ πππ 218Β° = 5 + 5
8
β΄ πππ 18Β° = 5 + 5
8 =
2(5 + 5)
16 =
10 + 2 5
4.
9. Find the value of π ππ 18Β°.
Hence find π ππ 36Β°.
Solution: Let π = 18Β°. Then 5π = 90Β°.
β΄ 3π + 2π = 90Β° or 2π = 90Β° β 3π.
β΄ π ππ 2π = π ππ (90Β° β 3π) = πππ 3π.
β΄ 2π ππ π. πππ π = 4πππ 3 π β 3πππ π.
Dividing by πππ π, we get, 2π ππ π = 4πππ 2 π β 3
β 2π ππ π = 4(1 β π ππ2 π) β 3
β 2π ππ π = 4 β 4π ππ2 π β 3
β 2π ππ π = 1 β 4π ππ2 π
β 4π ππ2 π + 2π ππ π β 1 = 0
β π ππ π = β 2 Β± 4 β 4(4)(β 1)
2 Γ 4 =
β 2 Β± 4 + 16
8 =
β 2 Β± 20
8
= β 2 Β± 2 5
8 =
2(β 1 Β± 5)
8 =
β 1 Β± 5
4.
Since π = 18Β°, π ππ π = π ππ 18Β° is positive. β΄ π ππ 18Β° = 5 β 1
4.
By using πππ 2π΄ = 1 β 2π ππ2π΄, we get,
πππ 36Β° = 1 β 2π ππ2 18Β° = 1 β 2 5 β 1
4
2
= 1 β 2 5 + 1 β 2 5
16 = 1 β
6 β 2 5
8 =
8 β (6 β 2 5)
8
= 2 + 2 5
8 =
1 + 5
4.
β΄ πππ 36Β° = 5 + 1
4.
β΄ π ππ2 36Β° = 1 β πππ 2 36Β° = 1 β 5 + 1
4
2
= 1 β 5 + 1 + 2 5
16
= 1 β 6 + 2 5
16 =
16 β 6 β 2 5
16 =
10 β 2 5
16
β΄ π ππ 36Β° = 10 β 2 5
4.
10. Show that πππ π΄ + πππ 120Β° + π΄ + πππ 120Β° β π΄ = 0.
Solution: We have
πππ 120Β° + π΄ = πππ 120Β°. πππ π΄ β π ππ 120Β°. π ππ π΄ and
πππ 120Β° β π΄ = πππ 120Β°. πππ π΄ + π ππ 120Β°. π ππ π΄
β΄ πππ 120Β° + π΄ + πππ 120Β° β π΄ = 2 πππ 120Β° πππ π΄
= 2 πππ 180Β° β 60Β° πππ π΄
= β 2πππ 60Β° . πππ π΄ = β 2 1
2 πππ π΄ = β πππ π΄.
β΄ πππ π΄ + πππ 120Β° + π΄ + πππ 120Β° β π΄
= πππ π΄ β πππ π΄ = 0 = π π»π.
11. Prove that πππ 20Β° πππ 40Β° πππ 60Β° πππ 80Β° = 1
16.
Solution: πΏπ»π =1
2 πππ 20Β° + 40Β° + πππ 20Β° β 40Β°
1
2. πππ 80Β°.
=1
4 πππ 60Β° + πππ β 20Β° . πππ 80Β°
=1
4
1
2 + πππ 20Β° . πππ 80Β° =
1
4
1 + 2 πππ 20Β°
2 . πππ 80Β°
=1
8 πππ 80Β° + 2 πππ 80Β° πππ 20Β°
=1
8 πππ 80Β° + πππ 80Β° + 20Β° + πππ 80Β° β 20Β°
=1
8 πππ 80Β° + πππ 100Β° + πππ 60Β°
= 1
8 πππ 80Β° + πππ 180Β° β 80Β° +
1
2
=1
8 πππ 80Β° β πππ 80Β° +
1
2 =
1
16
II method:
πππ 20Β° πππ 40Β° πππ 60Β° πππ 80Β°
= πππ 60Β° . πππ 20Β° πππ (60Β° β 20Β°) πππ (60Β° + 20Β°)
= πππ 60Β° . πππ 20Β° πππ 2 60Β° β π ππ2 20Β°
=1
2. πππ 20Β°
1
2
2 β ( 1 β πππ 2 20Β°)
=1
2. πππ 20Β°
1
4 β 1 + πππ 2 20Β°
=1
2. πππ 20Β° πππ 2 20Β° β
3
4
=1
2. πππ 20Β°
4πππ 2 20Β° β 3
4
=1
2.
4πππ 3 20Β° β 3 πππ 20Β°
4
= 1
2.
1
4. πππ (3 Γ 20Β°) =
1
2.
1
4. πππ 60Β° =
1
8.
1
2 =
1
16.
12. Show that π ππ π =π ππ 3π
1 + 2πππ 2π and hence prove that
π ππ15Β° = 3 β 1
2 2.
Solution: π ππ 3π
1 + 2πππ 2π =
3π ππ π β 4π ππ 3 π
1 + 2(1 β 2π ππ 2 π) =
π ππ π(3 β 4π ππ 2 π)
3 β 4π ππ 2 π = π ππ π.
β΄ π ππ π =π ππ 3π
1 + 2πππ 2π.
By taking π = 15Β° we get,
π ππ 15Β° =π ππ 3(15Β°)
1 + 2πππ 2(15Β°) =
π ππ 45Β°
1 + 2πππ 30Β° =
1 2
1 + 2 3 2 =
1
2 1 + 3
= 1
2 3 + 1 =
1
2 3 + 1 . 3 β 1
3 β 1 =
3 β 1
2 3 β 1 =
3 β 1
2 2.
13. Show that πππ 7π₯ β πππ 5π₯ + πππ 3π₯ β πππ π₯
π ππ 7π₯ β π ππ 5π₯ β π ππ 3π₯ + π ππ π₯ = πππ‘ 2π₯.
Solution: πππ 7π₯ β πππ 5π₯ + πππ 3π₯ β πππ π₯
π ππ 7π₯ β π ππ 5π₯ β π ππ 3π₯ + π ππ π₯ =
πππ 7π₯ + πππ 3π₯ β πππ 5π₯ β πππ π₯
π ππ 7π₯ β π ππ 3π₯ β π ππ 5π₯ + π ππ π₯
=πππ 7π₯ + πππ 3π₯ β (πππ 5π₯ + πππ π₯)
π ππ 7π₯ β π ππ 3π₯ β (π ππ 5π₯ β π ππ π₯)
=2 πππ
7π₯ + 3π₯2
.πππ 7π₯ β 3π₯
2 β 2 πππ 5π₯ + π₯
2 .πππ
5π₯ β π₯
2
2 πππ 7π₯ + 3π₯
2 .π ππ
7π₯ β 3π₯
2 β 2 πππ
5π₯ + π₯2
.π ππ 5π₯ β π₯
2
=2πππ 5π₯βπππ 2π₯ β 2πππ 3π₯βπππ 2π₯
2πππ 5π₯βπ ππ 2π₯ β 2πππ 3π₯βπ ππ 2π₯
=2πππ 2π₯(πππ 5π₯ β πππ 3π₯)
2π ππ 2π₯(πππ 5π₯ β πππ 3π₯) =
πππ 2π₯
π ππ 2π₯ = πππ‘ 2π₯.
14. Prove that πππ π₯ + πππ π¦ 2 + π ππ π₯ β π ππ π¦ 2 = 4πππ 2 π₯ + π¦
2 .
Solution: πππ π₯ + πππ π¦ 2 + π ππ π₯ β π ππ π¦ 2
= πππ 2π₯ + πππ 2 π¦ + 2πππ π₯. πππ π¦ + π ππ2 π₯ + π ππ2 π¦ β 2π ππ π₯. π ππ π¦
= πππ 2π₯ + π ππ2 π₯ + πππ 2 π¦ + π ππ2 π¦ + 2(πππ π₯. πππ π¦ β π ππ π₯. π ππ π¦)
= 1 + 1 + 2πππ (π₯ + π¦)
= 2 + 2πππ (π₯ + π¦) = 2 1 + πππ (π₯ + π¦) = 2.2πππ 2 π₯ + π¦
2
= 4πππ 2 π₯ + π¦
2 .
15. If π‘ππ π₯ = β 4
3,
π
2 < π₯ < π find π ππ
π₯
2, πππ
π₯
2 and π‘ππ
π₯
2.
Solution: Let πππ = 4 and πππ = 3. Then βπ¦π = 5.
Since π₯ is an angle lying in the II quadrant, πππ π₯ is negative.
β΄ πππ π₯ = β πππ
βπ¦π = β
4
5.
Now 2π ππ2 π₯
2 = 1 β πππ π₯ = 1 β β
4
5 =
5 + 4
5 =
9
5.
β΄ π ππ2 π₯
2 =
9
10. β΄ π ππ
π₯
2 = Β±
3
10.
Since π
2 < π₯ < π we get,
π
4 <
π₯
2 <
π
2.
β΄ π ππ π₯
2 > 0. β΄ π ππ
π₯
2 =
3
10.
Now 2πππ 2 π₯
2 = 1 + πππ π₯ = 1 + β
4
5 =
5 β 4
5 =
1
5.
β΄ πππ 2 π₯
2 =
1
10. β΄ πππ
π₯
2 = Β±
1
10.
Since π
2 < π₯ < π we get,
π
4 <
π₯
2 <
π
2.
β΄ πππ π₯
2 > 0. β΄ πππ
π₯
2 =
1
10.
Also π‘ππ π₯
2 =
π ππ π₯
2
πππ π₯
2
= 3 10
1 10 = 3.
16. Prove that π ππ2 π₯ + π ππ2 π₯ + 2π
3 + π ππ2 π₯ β
2π
3 =
3
2.
Solution: π ππ2 π₯ + π ππ2 π₯ + 2π
3 + π ππ2 π₯ β
2π
3
= 1 β πππ 2π₯
2 +
1 β πππ 2 π₯ + 2π
3
2 +
1 β πππ 2 π₯ β 2π
3
2
= 1 β πππ 2π₯
2 +
1 β πππ 2π₯ + 4π
3
2 +
1 β πππ 2π₯ β 4π
3
2
= 1 β πππ 2π₯ + 1 β πππ 2π₯ +
4π
3 + 1 β πππ 2π₯ β
4π
3
2
= 3 β πππ 2π₯ + πππ 2π₯ +
4π
3 + πππ 2π₯ β
4π
3
2
= 3 β πππ 2π₯ + 2 πππ 2π₯ .πππ
4π
3
2
{ πππ 4π
3 = πππ
3π + π
3 = πππ π +
π
3 = β πππ
π
3 = β
1
2 }
= 3 β πππ 2π₯ + 2 πππ 2π₯ . β
1
2
2 =
3 β πππ 2π₯ β πππ 2π₯
2 =
3
2.