ch-03 trigonometric functions
TRANSCRIPT
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Class XI Chapter 3 Trigonometric Functions Maths
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Exercise 3.1
Question 1:
Find the radian measures corresponding to the following degree measures:
(i) 25 (ii) 47 30' (iii) 240 (iv) 520
Answer
(i) 25
We know that 180 = radian
(ii) 47 30'
47 30' = degree [1 = 60']
degree
Since 180 = radian
(iii) 240We know that 180 = radian
(iv) 520
We know that 180 = radian
Question 2:
Find the degree measures corresponding to the following radian measures
.
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Class XI Chapter 3 Trigonometric Functions Maths
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(i) (ii) 4 (iii) (iv)
Answer
(i)
We know that radian = 180
(ii) 4
We know that radian = 180
(iii)
We know that radian = 180
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Class XI Chapter 3 Trigonometric Functions Maths
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(iv)
We know that radian = 180
Question 3:
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in
one second?
Answer
Number of revolutions made by the wheel in 1 minute = 360
Number of revolutions made by the wheel in 1 second =
In one complete revolution, the wheel turns an angle of 2 radian.
Hence, in 6 complete revolutions, it will turn an angle of 6 2 radian, i.e.,
12 radian
Thus, in one second, the wheel turns an angle of 12 radian.
Question 4:
Find the degree measure of the angle subtended at the centre of a circle of radius 100
cm by an arc of length 22 cm .
Answer
We know that in a circle of radius runit, if an arc of length lunit subtends an angle
radian at the centre, then
Therefore, forr = 100 cm, l = 22 cm, we have
Thus, the required angle is 1236.
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Class XI Chapter 3 Trigonometric Functions Maths
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Question 5:
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor
arc of the chord.
Answer
Diameter of the circle = 40 cm
Radius (r) of the circle =
Let AB be a chord (length = 20 cm) of the circle.
In OAB, OA = OB = Radius of circle = 20 cm
Also, AB = 20 cm
Thus, OAB is an equilateral triangle.
= 60 =
We know that in a circle of radius runit, if an arc of length lunit subtends an angle
radian at the centre, then .
Thus, the length of the minor arc of the chord is .
Question 6:
If in two circles, arcs of the same length subtend angles 60 and 75 at the centre, find
the ratio of their radii.
Answer
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Class XI Chapter 3 Trigonometric Functions Maths
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Let the radii of the two circles be and . Let an arc of length lsubtend an angle of 60
at the centre of the circle of radius r1, while let an arc of length l subtend an angle of 75
at the centre of the circle of radius r2.
Now, 60 = and 75 =
We know that in a circle of radius runit, if an arc of length lunit subtends an angle
radian at the centre, then .
Thus, the ratio of the radii is 5:4.
Question 7:
Find the angle in radian though which a pendulum swings if its length is 75 cm and the
tip describes an arc of length
(i) 10 cm (ii) 15 cm (iii) 21 cm
Answer
We know that in a circle of radius runit, if an arc of length lunit subtends an angle
radian at the centre, then .
It is given that r= 75 cm
(i) Here, l= 10 cm
(ii) Here, l = 15 cm
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Class XI Chapter 3 Trigonometric Functions Maths
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(iii) Here, l = 21 cm
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Exercise 3.2
Question 1:
Find the values of other five trigonometric functions if ,xlies in third
quadrant.
Answer
Sincexlies in the 3rdquadrant, the value of sinxwill be negative.
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Class XI Chapter 3 Trigonometric Functions Maths
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Question 2:
Find the values of other five trigonometric functions if ,xlies in second
quadrant.
Answer
Sincexlies in the 2ndquadrant, the value of cosxwill be negative
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Class XI Chapter 3 Trigonometric Functions Maths
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Question 3:
Find the values of other five trigonometric functions if ,xlies in third quadrant.
Answer
Sincex lies in the 3rdquadrant, the value of secxwill be negative.
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Class XI Chapter 3 Trigonometric Functions Maths
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Question 4:
Find the values of other five trigonometric functions if ,xlies in fourth
quadrant.
Answer
Sincexlies in the 4thquadrant, the value of sinxwill be negative.
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Class XI Chapter 3 Trigonometric Functions Maths
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Question 5:
Find the values of other five trigonometric functions if ,xlies in second
quadrant.
Answer
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Class XI Chapter 3 Trigonometric Functions Maths
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Sincexlies in the 2ndquadrant, the value of secxwill be negative.
secx=
Question 6:
Find the value of the trigonometric function sin 765
Answer
It is known that the values of sinx repeat after an interval of 2 or 360.
Question 7:
Find the value of the trigonometric function cosec (1410)
Answer
It is known that the values of cosecx repeat after an interval of 2 or 360.
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Class XI Chapter 3 Trigonometric Functions Maths
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Find the value of the trigonometric function
Answer
It is known that the values of tanx repeat after an interval of or 180.
Question 9:
Find the value of the trigonometric function
Answer
It is known that the values of sinx repeat after an interval of 2 or 360.
Question 10:
Find the value of the trigonometric function
Answer
It is known that the values of cotx repeat after an interval of or 180.
Question 8:
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Class XI Chapter 3 Trigonometric Functions Maths
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Exercise 3.3
Question 1:
Answer
L.H.S. =
Question 2:
Prove that
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Class XI Chapter 3 Trigonometric Functions Maths
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Answer
L.H.S. =
Question 3:
Prove that
Answer
L.H.S. =
Question 4:
Prove that
Answer
L.H.S =
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Class XI Chapter 3 Trigonometric Functions Maths
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Question 5:
Find the value of:
(i) sin 75
(ii) tan 15
Answer
(i) sin 75 = sin (45 + 30)
= sin 45 cos 30 + cos 45 sin 30
[sin (x+ y) = sinxcos y+ cosxsin y]
(ii) tan 15 = tan (45 30)
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Class XI Chapter 3 Trigonometric Functions Maths
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Question 6:
Prove that:
Answer
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Class XI Chapter 3 Trigonometric Functions Maths
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Question 7:
Prove that:
Answer
It is known that
L.H.S. =
Question 8:
Prove that
Answer
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Cl XI Ch t 3 T i t i F ti M th
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Answer
L.H.S. =
Question 10:
Prove that sin (n+ 1)xsin (n+ 2)x+ cos (n+ 1)xcos (n+ 2)x= cosxAnswer
L.H.S. = sin (n+ 1)xsin(n+ 2)x+ cos (n+ 1)xcos(n + 2)x
Question 11:
Prove that
Question 9:
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It is known that .
L.H.S. =
Question 12:
Prove that sin26x sin24x= sin 2xsin 10x
Answer
It is known that
L.H.S. = sin26x sin24x
= (sin 6x+ sin 4x) (sin 6x sin 4x)
= (2 sin 5xcosx) (2 cos 5xsinx)
= (2 sin 5xcos 5x) (2 sinxcosx)
= sin 10xsin 2x
Answer
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Class XI Chapter 3 Trigonometric Functions Maths
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Class XI Chapter 3 Trigonometric Functions Maths
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= R.H.S.
Question 13:
Prove that cos22x cos26x= sin 4x sin 8x
Answer
It is known that
L.H.S. = cos22x cos26x
= (cos 2x+ cos 6x) (cos 2x 6x)
= [2 cos 4xcos 2x] [2 sin 4x (sin 2x)]
= (2 sin 4xcos 4x) (2 sin 2xcos 2x)
= sin 8xsin 4x
= R.H.S.
Question 14:
Prove that sin 2x+ 2sin 4x+ sin 6x= 4cos2
xsin 4xAnswer
L.H.S. = sin 2x+ 2 sin 4x+ sin 6x
= [sin 2x+ sin 6x] + 2 sin 4x
= 2 sin 4xcos ( 2x) + 2 sin 4x
= 2 sin 4xcos 2x+ 2 sin 4x
= 2 sin 4x(cos 2x+ 1)
= 2 sin 4x(2 cos2
x 1 + 1)
= 2 sin 4x(2 cos2x)
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= 4cos2xsin 4x
= R.H.S.
Question 15:
Prove that cot 4x(sin 5x+ sin 3x) = cotx(sin 5x sin 3x)
Answer
L.H.S = cot 4x(sin 5x+ sin 3x)
= 2 cos 4xcosx
R.H.S. = cotx(sin 5x sin 3x)
= 2 cos 4x. cosx
L.H.S. = R.H.S.
Question 16:
Prove that
Answer
It is known that
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p g
Page 23 of 44
L.H.S =
Question 17:
Prove that
Answer
It is known that
L.H.S. =
Question 18:
Prove that
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Answer
It is known that
L.H.S. =
Question 19:
Prove that
Answer
It is known that
L.H.S. =
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Question 20:
Prove that
Answer
It is known that
L.H.S. =
Question 21:
Prove that
Answer
L.H.S. =
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Question 22:
Prove that cotxcot 2x cot 2xcot 3x cot 3xcotx= 1
Answer
L.H.S. = cotxcot 2x cot 2xcot 3x cot 3xcotx
= cotxcot 2x cot 3x(cot 2x+ cotx)
= cotxcot 2x cot (2x +x) (cot 2x+ cotx)
= cotx cot 2x (cot 2x cotx 1)
= 1 = R.H.S.
Question 23:
Prove that
Answer
It is known that .
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L.H.S. = tan 4x = tan 2(2x)
Question 24:
Prove that cos 4x= 1 8sin2x cos2x
AnswerL.H.S. = cos 4x
= cos 2(2x)
= 1 2 sin22x[cos 2A= 1 2 sin2A]
= 1 2(2 sinx cosx)2[sin2A= 2sinAcosA]
= 1 8 sin2xcos2x
= R.H.S.
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Question 25:
Prove that: cos 6x= 32 cos6x 48 cos4x+ 18 cos2x 1
Answer
L.H.S. = cos 6x
= cos 3(2x)
= 4 cos32x 3 cos2x[cos 3A= 4 cos3A 3 cosA]
= 4 [(2 cos2x 1)3 3 (2 cos2x 1) [cos 2x= 2 cos2x 1]
= 4 [(2 cos2x)3 (1)3 3 (2 cos2x)2+ 3 (2 cos2x)] 6cos2x+ 3
= 4 [8cos6x 1 12 cos4x+ 6 cos2x] 6 cos2x+ 3
= 32 cos6x 4 48 cos4x+ 24 cos2x 6 cos2x+ 3
= 32 cos6x 48 cos4x+ 18 cos2x 1
= R.H.S.
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Exercise 3.4
Question 1:
Find the principal and general solutions of the equation
Answer
Therefore, the principal solutions arex= and .
Therefore, the general solution is
Question 2:
Find the principal and general solutions of the equation
Answer
Therefore, the principal solutions arex= and .
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Therefore, the general solution is , where nZ
Question 3:
Find the principal and general solutions of the equation
Answer
Therefore, the principal solutions arex= and .
Therefore, the general solution is
Question 4:
Find the general solution of cosecx= 2
Answer
cosec x = 2
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Therefore, the principal solutions arex= .
Therefore, the general solution is
Question 5:
Find the general solution of the equation
Answer
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Question 6:
Find the general solution of the equation
Answer
Question 7:
Find the general solution of the equation
Answer
Therefore, the general solution is .
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Question 8:
Find the general solution of the equation
Answer
Therefore, the general solution is .
Question 9:
Find the general solution of the equation
Answer
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Therefore, the general solution is
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Class XI Chapter 3 Trigonometric Functions Maths
NCERT Miscellaneous Solution
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NCERT Miscellaneous Solution
Question 1: Prove that:
Answer
L.H.S.
= 0 = R.H.S
Question 2:
Prove that: (sin 3x + sinx) sinx + (cos 3x cosx) cosx = 0
Answer
L.H.S.
= (sin 3x + sinx) sinx + (cos 3x cosx) cosx
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= RH.S.
Question 3:
Prove that:
Answer
L.H.S. =
Question 4:
Prove that:
Answer
L.H.S. =
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Question 5:
Prove that:
Answer
It is known that .
L.H.S. =
Question 6:
Prove that:
Answer
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Class XI Chapter 3 Trigonometric Functions Maths
It is known that
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.
L.H.S. =
= tan 6x= R.H.S.
Question 7:
Prove that:
Answer
L.H.S. =
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Asxis in quadrant II, cosxis negative.
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Thus, the respective values of are .
Question 9:
Find for ,xin quadrant III
Answer
Here,xis in quadrant III.
Therefore, and are negative, whereas is positive.
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Now,
Thus, the respective values of are .
Question 10:
Find for ,xin quadrant II
Answer
Here,xis in quadrant II.
Therefore, , and are all positive.
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[cosxis negative in quadrant II]
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Thus, the respective values of are .
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