Download cos + x

Practice exercises - Exam 3 (Chapters 5 and 6)
1. Prove the following identities:
(a) sec θ − cos θ =
(b) cos
!
π
2
"
sin2 θ
cos θ
+ x = − sin(x)
(c) cot(A) =
sin(2A)
1−cos(2A)
2. Suppose sin A = − 35 with A in QIV and sin B =
12
13
with B in QII. Find the exact value of
sin(A + B).
3. Looking for familiar patterns, write the following expressions as a single trigonometric
function:
(a) cos(3x) cos(2) + sin(3x) sin(2)
(b) 2 sin(5t) cos(5t)
!
4. Find the exact value of cos 2 · sin−1
! ""
1
4
5. Prove that the equation sin2 x − cos2 x = 1 is ✿✿✿
not an identity by finding a counterexample.
6. Find all solutions in the interval 0 ≤ x < 2π to the equation
√
3 tan x − 1 = 0
7. Approximate to the nearest tenth of a degree all solutions in the interval 0◦ ≤ θ < 360◦ to
the equation 3 sin2 θ + 11 sin θ − 4 = 0
8. Find all degree solutions to the equation sin(2θ) − sin(θ) = 0
9. Give an example of a trigonometric equation with no solutions.
10. Find all radian solutions the the equation sin(3x) cos(x) + cos(3x) sin(x) = −1
Answers
1. Transform one side into the other. Do not work on both sides.
(a) sec θ − cos θ =
(b) cos
(c)
!
π
2
"
1
cos θ
+ x = cos
sin(2A)
1−cos(2A)
=
− cos θ =
! "
π
2
1
cos θ
cos x − sin
2 sin A cos A
1−(1−2 sin2 A)
=
cos2 θ
cos θ
−
! "
π
2
2 sin A cos A
2 sin2 A
=
1−cos2 θ
cos θ
=
sin2 θ
cos θ
sin x = 0 · cos x − 1 · sin x = 0 − sin x = − sin x
=
sin A·cos A
sin A·sin A
cos A
sin A
=
= cot A
4
and cos B =
5 "
! "! "
5
− 13 + 45 12
13
2. Draw a picture of A in QIV and B in QII to find cos A =
!
Then sin(A + B) = sin A cos B + cos A sin B = − 35
"!
5
− 13
.
=
63
65
3. The first expression comes from a difference formula, the second from a double angle
formula:
(a) cos(3x − 2)
(b) sin(10t)
4. Thinking of sin−1
angle θ,
! "
1
4
as an angle θ, we have cos(2θ). Drawing a right triangle with the
cos(2θ) = cos2 θ − sin2 θ =
! √ "2
15
4
−
! "2
1
4
=
14
16
=
7
8
5. One possible counterexample is x = 0.
For x = 0, the left hand side is 02 − 12 = −1 and −1 ̸= 1.
6. For tan x =
√1
3
we get tan x̂ =
√1
3
so that the reference angle x̂ =
Tangent is positive in QI and QIII giving x =
7. Factoring gives sin θ =
1
3
π
6
or x =
π
6
7π
6
or sin θ = −4.
The second equation has no solutions, and from sin θ̂ =
1
3
we find that the reference angle
θ̂ ≈ 19.5◦ .
Sine is positive in QI and QII giving θ ≈ 19.5◦ or θ ≈ 160.5◦
8. Using a double angle formula, 2 sin θ cos θ − sin θ = 0.
Factoring out the sine gives sin θ = 0 or cos θ =
1
2
From these equations, θ = k · 180◦ or θ = 60◦ + k · 360◦ or θ = 300◦ + k · 360◦
9. There are many possible equations; one possibility is cos x = 2.
10. The left hand side is a sum formula for sin(3x + x).
We have sin(4x) = −1 or, substituting t = 4x, sin(t) = −1
To make the sine equal to −1, the angle t must be of the form t =
Since t = 4x, dividing by four gives the solutions x =
3π
8
+k·
π
2
3π
2
+ k · 2π