trigonometric equations reminders i) radians converting between degrees and radians:
TRANSCRIPT
Trigonometric Equations
Reminders
i) Radians 180 radians o
Converting between degrees and radians:
120 120 x ??? radianso 120 121 00.8
o 2
3 radians
5 5
6 6 x ??? degees
5 5.
6 6
180
150o
ii) Exact Values45o right-angled triangle:
1
1
2
45o
1cos45 sin 45
2o o
tan 45 1o
Equilateral triangle:
3sin60
2o
1
2 2
1
60o
30o
3
1cos60
2o
1sin30
2o
3cos30
2o
tan60 3o 1
tan303
o
degrees 0o 30o 45o 60o 90o
radians
sin
cos
tan
0 6
3
4
3
2
0
1
0
1
23
2
1
21
3
2
1
2
1
20
1
31
Example: What is the exact value of sin 240o ?
240 180 60 sin(180 ) sin 3sin 240
2o
iii) Trigonometric Graphs:
sin oy xy
x
Amplitude
Period
360o0
-1
1
y
x
Period = 360o
Amplitude = 1
cos oy xy
x
Period
Amplitude
360o
-1
0
1
y
x
Period = 360o
Amplitude = 1
tan oy xy
x
Period
0o
360o270o
180o
90o
y
x
Period = 180o
Amplitude cannot be defined.
Solving Trigonometric Equations
Example: 2cos3 1 0 (0 360 )Solve ox x
Step 1: Re-Arrange 2cos3 1 0
2cos
1cos
3
3
1
2x
x
x
Step 2: consider what solutions are expected
C
AS
T
0o180o
270o
90o
3
2
2
All Sch…Talk Cr*!p
1cos3
2x
cas
t
cos 3x is positive so solutions in the first and fourth quadrants
0 360 2Since has solutionsox
0 3 1080 6Then has solutionsox
x 3 x 3
Step 3: Solve the equation1
cos32
x
1 13 cos2
x
3x = 60o 300o 420o 660o 780o 1020o
60 (360-60) (360+60) (720-60) (720+60) (1080-60)
1st quadrant4th quadrant cos wave repeats every 360o
x = 20o 100o 130o 220o 260o 340o
Example: 1 2 sin6 0 (0 180 )Solve ot t
Step 1: Re-Arrange1
sin62
t
Step 2: consider what solutions are expected
cas
tsin 6t is negative so solutions in the third and fourth quadrants
0 180 2Since has solutionsot
0 6 1080 12Then has solutionsot
x 6 x 6
Step 3: Solve the equation
1 16 sin
2t
6t = 225o 315o 585o 675o 945o 1035o
180+45 (360-45) (360+180+45) (720-45) (720+180+45) (1080-45)
3rd quadrant 4th quadrant sin wave repeats every 360o
t = 39.1o 52.5o 97.5o 112.5o 157.5o 172.5o
1sin6
2t
1 1sin 45
2st (1 quadrant)o
Example: 2sin(2 ) 1 (0 2 )3
Solve x x
The solution is to be in radians – but work in degrees and convert at the end.
Step 1: Re-Arrange 1sin(2 60 )
2ox
Step 2: consider what solutions are expected
cas
tsin (2x – 60o ) is positive so solutions in the first and second quadrants
0 360 2Since has solutionsox
0 2 720 4Then has solutionsox
x 2 x 2
Step 3: Solve the equation 1sin(2 60 )
2ox
1 12 60 sin2
ox
1 1sin 302
o
st (1 quadrant)
2x = 90o or 210o
2x-60 = 0 + 30 or (180 - 30))
1st quadrant 2nd quadrant
Now Add on the period of the wave to each of the values found in the first wave. i.e.
x = 45o + 180 or (105 + 180)
4 3
3 2 3 2x
X = 45 or 105
Harder Example: 2tan 3 (0 2 )Solve x x
Step 1: Re-Arrange tan 3x
Step 2: Consider what solutions are expected
We need to solve 2 equations.y
x
0o360o
270o
180o
90o
y
x
tan 3x
tan 3x
Expect 2 +ve solutions
Expect 2 -ve solutions
1tan 3 (603
st in the 1 quadrant)o
Step 3: Solve the equation tan 3x
1tan 3 (603
st in the 1 quadrant)o
tan 3x3 3
and x x
tan 3x 23 3
and x x
2 4 5, , ,3 3 3 3
x x x x
Harder Example: 23sin 4sin 1 0 (0 360 )Solve ox x x
Step 1: Re-Arrange 23 4 1 0Cp. w. p p
(3sin 1)(sin 1) 0x x
Step 2: Consider what solutions are expected
y
x360o0
-1
1
y
x
We need to solve 2 equations.
sin 1x1
sin3
x
Just ONE solution
Two solutions
Step 3: Solve the equations sin 1x 1sin
3x
1 1sin 19.53
o
In the 1st quadrant
1 1sin 19.5 180 19.53
or o ox
x = 19.5o , 90o , 160.5o
Even Harder Example: 25sin 2 2cos (0 2 )Solve x x x
Step 1: Re-Arrange Remember this ????2 2
2 2
2 2
sin cos 1
cos 1 sin
sin 1 cos
25(1 cos ) 2 2cosx x
23 2cos 5cos 0x x
(3 5cos )(1 cos ) 0x x
Step 2: Consider what solutions are expected
We need to solve 2 equations. 3cos
5x
cos 1x Just ONE solution
Two solutions
1 3cos 0.935
radiansx
0.93, , 2 0.93 radiansx
Step 3: Solve the equations
cas
t