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Trigonometry II
Trigonometry II©
Standards
ANSWERS
Copyright
2006,Test
Barry
Mabillard.
0
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2
4
1. If cos α = − , where sin α > 0 , and sin β = , where tan β > 0 , determine the exact
5
9
value of cos (α + β )
First determine the quadrant α is in:
cos α < 0
sin α > 0
Then determine the quadrant β is in:
sin β > 0
tan β > 0
Now solve the triangles for both α and β; use Pythagoras to find the unknown sides.
a 2 + b2 = c2
a 2 + b2 = c 2
( −2 )
( 4)
2
+ b 2 = 52
2
+ b 2 = 92
4 + b 2 = 25
16 + b 2 = 81
b 2 = 21
b 2 = 65
b = 21
b = 65
At this point, state all required trigonometric ratios:
2
65
cos β =
cos α = −
9
5
4
21
sin β =
sin α =
9
5
Finally, evaluate cos (α + β )
cos (α + β ) = cos α cos β − sin α sin β
⎛ 2 ⎞ ⎛ 65 ⎞ ⎛ 21 ⎞ ⎛ 4 ⎞
cos (α + β ) = ⎜ − ⎟ ⎜⎜
⎟−⎜
⎟⎜ ⎟
⎝ 5 ⎠ ⎝ 9 ⎠⎟ ⎝⎜ 5 ⎠⎟ ⎝ 9 ⎠
2 65 4 21
−
45
45
2 65 - 4 21
cos (α + β ) =
45
cos (α + β ) =
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2. Express 1 − sin 2 (α + β ) using only cosine.
Start with the identity: sin 2 θ + cos 2 θ = 1
Rearrange to get: cos 2 θ = 1 − sin 2 θ
It follows that: cos 2 (α + β ) = 1 − sin 2 (α + β )
The answer is cos 2 (α + β )
3. If cos 3θ =
3
, where 3θ is an acute angle, determine the exact value of θ.
2
3
π 11π
, you get x = ,
If you evaluate cosx =
.
6 6
2
π
is acute (<90°).
6
π
π 1
π
Divide by 3 to get the angle θ: ÷ 3 = × =
6
6 3 18
You do not want to use any other angles since only
3
in your calculator using
2
degree mode, then convert the answer to a radian fraction.
*Alternatively, you could graph each side of cos 3θ =
4. Determine the exact value of tan (15o )
Rewrite as tan (45 o - 30 o ) and use the formula tan (α − β ) =
tan 45o − tan 300
1 + tan 45o tan 300
1
1
1−
1−
3
3
=
=
1
⎛ 1 ⎞ 1+
1 + (1) ⎜
⎟
3
⎝ 3⎠
tan α − tan β
1 + tan α tan β
tan ( 45o − 300 ) =
=
3
−
3
3
+
3
1
3
1
3
=
3 −1
3
3 +1
3
=
3 −1
3
×
3
3 +1
=
3 -1
3 +1
Get a common denominator for top & bottom
Evaluate the required tan ratios before
doing the calculation on the left.
1
o
sin
30
1 2
1
tan 30o =
= 2 = ×
=
o
cos 30
3 2
3
3
2
3
o
sin 45
tan 45o =
= 2 =1
o
cos 45
3
2
Divide fractions by multiplying the reciprocal
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5. Prove the identity:
1 − cot 2 x
= sin 2 x − cos 2 x
1 + cot 2 x
1- cot 2 x
1+ cot 2 x
1- cot 2 x
=
csc 2 x
= (1- cot 2 x ) sin 2 x
⎛ cos 2 x ⎞
2
= ⎜1⎟ sin x
2
sin
x
⎝
⎠
2
= sin x - cos 2 x
6. Solve for x, where 0 ≤ x ≤ 2π : csc2 x − csc x = 2
Rewrite as csc 2 x - cscx - 2 = 0
Then factor: (cscx - 2 )(cscx + 1 ) = 0
Now equate each set of brackets to zero and solve for x.
cscx - 2 = 0
cscx = 2
1
2
π 5π
x= ,
6 6
sinx =
cscx + 1= 0
cscx = -1
sinx = -1
3π
x=
2
The answer is x =
π 5 π 3π
6
,
6
,
2
7. Solve for x: 2 cos 2 x + cos x − 1 = 0
Factor to obtain: (2cosx - 1 )(cosx + 1 ) = 0
Now equate each set of brackets to zero and solve for x.
2cosx - 1= 0
2cosx = 1
1
2
π 5π
x= ,
3 3
cosx =
cosx + 1= 0
cosx = -1
x=π
The answer is x =
Trigonometry II Standards Test ANSWERS
π
3
,π ,
3
5π
3
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8. Simplify:
sin x cot x + cos x
sin x
⎛ cosx ⎞
sinx ⎜
⎟+ cosx
sinx ⎠
⎝
=
sinx
cosx + cosx
=
sinx
2cosx
=
sinx
= 2cotx
9. Solve ( 2 cos θ + 1) ( tan θ − 1) = 0 for θ in the interval
π
≤ x ≤ 2π
2
Equate each set of brackets to zero and solve for θ.
2cosθ + 1= 0
2cosθ = -1
tanθ - 1= 0
tanθ = 1
π 5π
θ= ,
4 4
1
2
2 π 4π
θ=
,
3 3
cosθ = -
θ=
2 π 5 π 4π
,
,
3
4
3
(Omit
π
4
since it lies outside the specified domain.)
10. Determine the exact value of
11π
a) cos
12
11π
First convert
to degrees. This will make
12
b) sec
sec165 o =
the remainder of the calculation easier than
working in radian fractions.
(
)
11π 180 o
×
= 165 o
π
12
Write as cos 120 o + 45 o , then apply the
1
cos165 o
This is simply the reciprocal of
the answer you found in part a)
The answer is
cosine sum formula
(
11π
12
)
4
- 2- 6
o
o
o
o
o
o
cos 120 + 45
= cos120 cos45 - sin120 cos45
⎛ 1 ⎞⎛ 2 ⎞ ⎛ 3 ⎞⎛ 2 ⎞
= ⎜- ⎟ ⎜⎜
⎟- ⎜
⎟⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
- 2
6
4
4
- 2- 6
=
4
=
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11. Prove the identity
12. If tan θ = −
cot θ sec 2 θ
= tan θ
cot 2 θ + 1
cotθsec 2θ
cot 2θ + 1
cotθsec 2θ
=
csc 2θ
1
2
= cotθ • cos θ
1
sin 2θ
cosθ ⎛ 1 ⎞
2
=
⎜
⎟ ( sin θ )
sinθ ⎝ cos 2θ ⎠
sinθ
=
cosθ
= tanθ
3
3π
and
< θ < 2π , state the exact value of cos 2θ
2
2
Draw the triangle and find the
hypotenuse using Pythagoras:
a 2 + b2 = c2
( 2 ) + ( −3)
2
2
= c2
4 + 9 = c2
13 = c 2
c = 13
From the diagram, cosθ =
2
13
Now use the formula: cos2θ = 2cos 2θ - 1
2
⎛ 2 ⎞
cos2θ = 2 ⎜
⎟ -1
⎝ 13 ⎠
⎛ 4 ⎞
cos2θ = 2 ⎜
⎟- 1
⎝ 13 ⎠
8
cos2θ =
-1
13
5
cos2θ = 13
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3
2
To solve this equation algebraically, you need to determine
the solutions within 2 rotations of the unit circle.
3
π
2π
Solve the equation sinx =
to get the angles
,
and
2
3
3
both of which are within the first rotation.
13. Solve for θ over [ 0, 2π ] : sin 2θ =
Add the period to each one to get the co-terminal
angles in the second rotation.
π
π 6π 7 π
+ 2π = +
=
3
3
3
3
2π
2π 6π 8π
+ 2π =
+
=
3
3
3
3
Finally, take all your solutions
π 2π 7 π 8π
and divide by 2.
,
,
3
3
3
π 2π 7 π 8π
The answer is ,
,
,
6 6
6 6
3
,
14. Solve the following equation for x ∈ R : 2 cos 2θ + 1 = 0
Get cos2θ by itself before trying to determine the angles:
2cos2θ = -1
-1
cos2θ =
2
To solve this equation algebraically, you need to determine the
solutions within 2 rotations of the unit circle.
1
2π
4π
to get the angles
,
Solve the equation cosx = and
2
3
3
both of which are within the first rotation.
Add the period to each one to get the co-terminal
angles in the second rotation.
2π
2π 6π 8π
+ 2π =
+
=
3
3
3
3
4π
4 π 6 π 10 π
π
+ 2π =
+
=
3
3
3
3
2 π 4 π 8 π 10 π
and divide by 2.
,
,
,
3 3 3
3
2 π 4 π 8 π 10 π
,
,
,
The answer is
6 6 6
6
Finally, take all your solutions
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15. If α and β are second quadrant angles, and cos α = −
determine the exact value of sin (α − β )
2
1
and sin β = ,
3
4
Solve the triangles for both α and β; use Pythagoras to find the unknown sides.
a 2 + b2 = c2
a 2 + b2 = c 2
( −2 )
12 + b 2 = 42
+ b 2 = 32
2
4 + b2 = 9
1 + b 2 = 16
b =5
b 2 = 15
b= 5
b = 15
2
At this point, state all required trigonometric ratios:
2
3
5
sin α =
3
cos α = −
Use a negative since the b-value
lies on the negative x-axis.
− 15
4
1
sin β =
4
cos β =
Finally, evaluate sin (α - β )
sin (α − β ) = sin α cos β − cos α sin β
⎛ 5 ⎞ ⎛ − 15 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞
sin (α − β ) = ⎜⎜
⎟⎟ ⎜⎜
⎟⎟ − ⎜ − ⎟ ⎜ ⎟
⎝ 3 ⎠⎝ 4 ⎠ ⎝ 3 ⎠⎝ 4 ⎠
− 75 2
+
12
12
2 - 75
sin (α − β ) =
12
sin (α − β ) =
16. Solve for x over the interval [ 0, 2π ] for sin 2 x = sin x
Set the equation to zero: sin 2 x - sinx = 0
Then factor: sinx ( sinx - 1 ) = 0
Now set each factor equal to zero and solve for the angles
sinx = 0
x = 0, π ,2 π
sinx - 1= 0
sinx = 1
x=
π
2
The solution is
π
x=0, , π ,2π
2
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17. Prove
1
1
+
= 2 csc θ
1 + cos θ 1 − cos θ
1
1
⎛
⎞ ⎛ 1- cosθ ⎞ ⎛
⎞ ⎛ 1+ cosθ ⎞
⎜
⎟⎜
⎟+ ⎜
⎟⎜
⎟
⎝ 1+ cosθ ⎠ ⎝ 1- cosθ ⎠ ⎝ 1- cosθ ⎠ ⎝ 1+ cosθ ⎠
1
1
⎛
⎞ ⎛ 1- cosθ ⎞ ⎛
⎞ ⎛ 1+ cosθ ⎞
=⎜
⎟⎜
⎟+ ⎜
⎟⎜
⎟
⎝ 1+ cosθ ⎠ ⎝ 1- cosθ ⎠ ⎝ 1- cosθ ⎠ ⎝ 1+ cosθ ⎠
1- cosθ
1+ cosθ
=
+
2
1- cos θ 1- cos 2θ
1- cosθ + 1+ cosθ
=
1- cos 2θ
2
=
1- cos 2θ
2
=
sin 2θ
= 2cscθ
18. Solve for x over the interval [ 0, 2π ] : 1 + tan 2 x = 3 .
State the solutions as radians to three decimal places
First isolate tanx :
1+ tan 2 x = 3
tan 2 x = 2
tanx = ± 2
Now solve in your calculator by graphing
Y1 = tan x
Y2 = + 2
Y3 = − 2
The answer is x = 0.9555, 2.186, 4.097, 5.328
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19.
a) Prove
sin 2 x
= tan x
1 + cos 2 x
sin2x
1+ cos2x
2sinxcosx
=
1+ (2cos 2 x - 1 )
2sinxcosx
2cos 2 x
sinx
=
cosx
= tanx
=
sin 2 x
is undefined
1 + cos 2 x
The expression is undefined when 1+ cos2x = 0
1+ cos2x = 0
b) State a value of x where
cos2x = -1
First solve cosx = -1 over two rotations:
The solution to this is x = π ,3 π
Now divide by the 2 to get the solutions to cos2x = -1
π 3π
The final answer is x = ,
2 2
(You could also solve this by graphing Y1 = cos2x & Y2 = -1
in your calculator and finding the points of intersection)
⎛π ⎞
20. Find the exact value of 2 cos 2 ⎜ ⎟ − 1
⎝8⎠
Use the identity cos2α = 2cos 2 α - 1
⎛π ⎞
cos2α = 2cos 2 ⎜ ⎟- 1
⎝8 ⎠
What goes here is
double what goes here.
π
8
=
π
⎛π
= 2cos 2 ⎜
4
⎝8
π
2
The answer is: cos =
4
2
We now have the equation cos
Trigonometry II Standards Test ANSWERS
9
π
2×
4
⎞
⎟- 1
⎠
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⎛
cos 2 x ⎞
+
x
sin
⎜
⎟
sin x ⎠
21. Simplify ⎝
csc x
⎛
cos 2 x ⎞
sinx
+
⎜
⎟
sinx ⎠
⎝
cscx
⎛
cos 2 x
= ⎜ sinx +
sinx
⎝
⎞
⎟× sinx
⎠
= sin 2 x + cos 2 x
=1
22. Find the exact value of:
⎛ 7π ⎞
a) cos ⎜
⎟
⎝ 12 ⎠
First express the angle in terms of degrees:
7 π 180 o
×
= 105 o
π
12
cos (105 o ) = cos (60 o + 45 o )
cos (60 o + 45 o ) = cos60 o cos45 o - sin60 o sin45 o
⎛ 1 ⎞⎛ 2 ⎞ ⎛ 3 ⎞⎛ 2 ⎞
= ⎜ ⎟ ⎜⎜
⎟- ⎜
⎟⎜
⎟
⎝ 2 ⎠ ⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
2
6
4
4
2- 6
=
4
=
⎛ 7π ⎞
b) sec ⎜
⎟
⎝ 12 ⎠
This is simply the reciprocal of the answer you found in part a)
4
⎛7 π ⎞
sec ⎜
⎟=
2- 6
⎝ 12 ⎠
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23. Express tan 2 θ using only sin θ
sin 2θ
cos 2θ
Rewrite the identity sin 2θ + cos 2θ = 1 as cos 2θ = 1- sin 2θ
Start with the identity tan 2θ =
Finally,
24. Prove:
sin 2θ
sin 2θ
=
cos 2θ 1- sin 2θ
cos x
sin x + 1
+
= 2sec x
sin x + 1 cos x
⎛ cosx ⎞ ⎛ cosx ⎞ ⎛ sinx + 1 ⎞ ⎛ sinx + 1 ⎞
⎜
⎟⎜
⎟+ ⎜
⎟⎜
⎟
⎝ sinx + 1 ⎠ ⎝ cosx ⎠ ⎝ cosx ⎠ ⎝ sinx + 1 ⎠
cos 2 x
sin 2 x + 2sinx + 1
+
cosx ( sinx + 1 )
cosx ( sinx + 1 )
cos 2 x + sin 2 x + 2sinx + 1
cosx ( sinx + 1 )
1+ 2sinx + 1
cosx ( sinx + 1 )
2 + 2sinx
cosx ( sinx + 1 )
2 (1+ sinx )
cosx ( sinx + 1 )
2
cosx
= 2secx
=
25. Solve for θ in the interval [ 0, 2π ] : 2sin 2 θ − 3sin θ + 1 = 0
Factor to obtain: (2sinθ - 1 )( sinθ - 1 ) = 0
2sinθ - 1= 0
2sinθ = 1
1
sinθ =
2
π 5π
θ= ,
6 6
Now solve for the angles:
sinθ - 1= 0
sinθ = 1
θ=
π
2
The answer is θ =
Trigonometry II Standards Test ANSWERS
π π 5π
, ,
6 2 6
11
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26. If cos θ = −
4
and csc θ > 0 , determine the exact value of:
5
a) tan 2θ
First determine which quadrant
the angle is found in:
Now fill in the triangle
and use Pythagoras to
determine the unknown
side:
a 2 + b2 = c 2
( −4 )
2
+ b 2 = 52
16 + b 2 = 25
b2 = 9
b=3
Use the formula tan2θ =
⎛ 3
2 ⎜⎝ 4
tan2θ =
⎛ 3
1- ⎜⎝ 4
6
4
=
9
116
6
4
=
16 9
16 16
6
= 4
7
16
6 16
=- ×
4
7
24
=7
2tanθ
1- tan 2θ
⎞
⎟
⎠
2
⎞
⎟
⎠
From the triangle,
3
sinθ =
5
4
cosθ = 5
3
tanθ = 4
b) sin 2θ
Use the formula sin2θ = 2sinθcosθ
⎛ 3 ⎞⎛ 4 ⎞
sin2θ = 2 ⎜ ⎟⎜- ⎟
⎝ 5 ⎠⎝ 5 ⎠
24
sin2θ = 25
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27. Solve the equation csc θ = −3 , where θ ∈ R . State your solution to
three decimal places.
Solve by graphing:
The question specifies the domain as θ ∈ R , so
a general solution is required.
The (positive) solutions are 3.481 and 5.943
Since the solutions repeat themselves every period,
write the general solution as:
⎧3.481+ 2k π , k ∈ I ⎫
θ= ⎨
⎬
⎩5.943+ 2k π , k ∈ I ⎭
28. Show that sin 8x is equivalent to 2sin 4 x cos 4 x
Start with the identity sin (2x ) = 2sinxcosx
What goes here is
double what goes here.
In the equation sin8x = 2sin4xcos4x
the 4x here
becomes the
8x here
29. Express cot θ sec θ as a single trigonometric function
cotθsecθ
cosθ
1
=
×
sinθ cosθ
1
=
sinθ
= cscθ
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30. If sin θ =
12
and cos θ < 0 , find the exact value of
13
a) tan θ
First determine which quadrant
the angle is found in:
a2 + b2 = c2
(12 )
2
+ b 2 = 132
144 + b 2 = 169
b 2 = 65
b=5
From the triangle, the
trig rations are:
12
sinθ =
13
5
cosθ = 13
12
tanθ = 5
The answer is tanθ = -
π⎞
⎛
b) cos ⎜ θ − ⎟
4⎠
⎝
12
5
π⎞
⎛
Expand cos ⎜θ - ⎟ using the subtraction law for cosine:
⎝ 4⎠
π⎞
π
π
⎛
cos ⎜θ - ⎟ = cosθcos + sinθsin
4
4
⎝ 4⎠
Now use the values found in part a) from the triangle,
and values from the unit circle:
π
π ⎛ 5 ⎞ ⎛ 2 ⎞ ⎛ 12 ⎞ ⎛ 2 ⎞
cosθcos
4
+ sinθsin
4
= ⎜⎟
⎟+ ⎜
⎟⎜
⎟⎜
⎝ 13 ⎠ ⎜⎝ 2 ⎟⎠ ⎝ 13 ⎠ ⎜⎝ 2 ⎟⎠
5 2 12 2
+
26
26
12 2 - 5 2
=
26
=-
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31. Find the exact values of x in the interval [ 0, 2π ] : cos 2 x + cos x + 1 = sin 2 x
cos 2 x + cosx + 1= sin 2 x
cos 2 x + cosx = sin 2 x - 1
cos 2 x + cosx = -cos 2 x
2cos 2 x + cosx = 0
2cosx + 1= 0
2cosx = -1
1
cosx = 2
2 π 4π
x=
,
3 3
cosx (2cosx + 1 ) = 0
Now solve for x:
cosx = 0
π 3π
x= ,
2 2
The full solution set is x =
32. Prove the identity:
π 2 π 4π 3π
2
,
3
,
3
,
2
cos x − cos3 x
= cot x
sin 3 x
cosx - cos 3 x
sin 3 x
cosx (1- cos 2 x )
=
sin 3 x
cosxsin 2 x
=
sin 3 x
cosx
=
sinx
= cotx
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33. Given the point shown on the circle, determine the value of tan 2β
Recall that the x-coordinate is cosθ , and the y-coordinate is sinθ .
5
adj
=
13 hyp
12 opp
sinβ =
=
13 hyp
12
Therefore, tanβ =
5
cosβ =
Now use the identity tan2θ =
2tanθ
1- tan 2θ
⎛ 12 ⎞
2⎜
⎟
⎝ 5 ⎠
tan2θ =
2
⎛ 12 ⎞
1- ⎜
⎟
⎝ 5 ⎠
24
5
tan2θ =
144
125
24
5
tan2θ =
25 144
25 25
24
tan2θ = 5
119
−
25
24
25
tan2θ =
×5
119
120
tan2θ = 119
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