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Trigonometry for Calculus: Self Assessment
1. Find the exact value of the expression or say why it is undefined.
(a) sin
!!
(b) cos −
= 𝟏/𝟐
!
(e) sec 0 = 𝟏 !"
(f) tan
!
!
!"
(g) cos
= − 𝟑 !
!
(c) tan
=𝟎
!
= 𝟑 (d) sin −
𝟏
(h) csc −
= − !
𝟐
!"
!
!
!
=−
𝟐
𝟐
= − 2 2. Find the exact value of the expression or say why it is undefined.
!
!
(a) sin 𝑥2 𝜋 , 𝑥 =
(b) 𝑢cos ! 𝑢, 𝑢 =
!
𝜋
= sin
=
4
!
𝟐
𝟐
(c) cot 𝜋sin𝑎 , 𝑎 =
!
!
(d)
!
cos!
= cot 𝜋sin ! = cot 𝜋 =
sin!
,
!
= cos
!!
!
!
cos ! !
sin! !
!
= +
=
!
!
!
!
=
𝝅
𝟖
, 𝑥 = 20
cos ! ! ! sin! !
!
which is undefined, because sin𝝅 = 𝟎.
!
!
=
!
!
=
𝟏
𝟐𝟎
.
3. Find all solutions of the given equation on the indicated interval.
!
(a) 2cos𝑥 = 1, 0 ≤ 𝑥 ≤ 2𝜋
(b) sin 3𝜃 ! = , 0 ≤ 𝜃 ≤ 2𝜋
!
!
⇒ cos𝑥 =
!
(c) 𝑢cos
𝟓𝝅
𝟑
𝟑
.
⇒ no solutions, since sin𝑥 ≤ 1 for all 𝑥.
(c) 4sin2 𝑡 = 1, −
+ 𝑢 = 0, −4𝜋 ≤ 𝑢 ≤ −2𝜋
!
⇒ 𝑢 cos
𝝅
⇒ 𝒙 = or 𝒙 =
!
!
!
+ 1 = 0 ⇒ 𝑢 = 0 or cos
!
!
⇒ sin2 𝑡 =
= −1.
𝝅
!
the second, observe that −4𝜋 ≤ 𝑢 ≤ −2𝜋 ⇒ −𝜋 ≤
!
!
= −1 ⇒
!
!
sin!
= sin𝑥
= cos!
+
cos!
=
𝟏
cos𝒙
(b)
!
+ cos ! 𝑥
cos!
sin! !
cos ! !
= sec𝒙
!
!
!
!
𝝅
!
𝟔
!
sin!!
− cot𝑥sin𝑥
sin!
2sin!cos!
sin!
−
cos!
sin!
sin𝑥 = 2cos𝑥 − cos𝑥
= cos𝒙.
!
!
< 𝜃 < 𝜋. Find cos𝜃, tan𝜃, and sec𝜃.
Since sin 𝜃 + cos 𝜃 = 1, cos 𝜃 = 1 − 0.6! = 0.64. !
Therefore, cos𝜃 = ±0.8. Since cosine is negative for < 𝜃 < 𝜋, cos𝜽 = −𝟎. 𝟖.
Then tan𝜽 =
𝟎.𝟔
!𝟎.𝟖
!
≤𝑡≤
⇒ sin𝑡 = ± , so
𝟔
!
≤− .
=
cos!
5. (a) Suppose that sin𝜃 = 0.6 and
!
!
= −𝜋, or 𝒖 = −𝟒𝝅.
4. Simplify the given expression.
(a) sin𝑥tan𝑥 + cos ! 𝑥sec𝑥
!
!
𝒕 = − or 𝒕 = .
The first solution is not in the given interval. For
Then cos
!
!
= −𝟎. 𝟕𝟓 and sec𝜽 =
(b) Suppose that tan𝑥 = − 3 and
Since tan𝑥 = − 3 = −
!
!
Since cosine is positive for
and csc𝒙 =
𝟏
sin𝒙
=−
𝟐
𝟑.
!
!
!!
!
=
sin!
cos!
!!
!
𝟏
!𝟎.𝟖
!
= −𝟏. 𝟐𝟓.
< 𝑥 < 2𝜋. Find sin𝑥, cos𝑥, and csc𝑥.
, then sin𝑥 = ±
!
!
and cos𝑥 = ∓
< 𝑥 < 2𝜋, it must be that sin𝒙 = −
𝟑
𝟐
!
!
(the signs are opposite.)
, cos𝒙 =
𝟏
𝟐
Trigonometry for Calculus: Self Assessment
Notes on the Unit Circle Diagram
* The circle has radius 1 and points 𝑥, 𝑦 on the circle satisfy the equation 𝒙𝟐 + 𝒚𝟐 = 𝟏.
* Angles are displayed inside the circle (in degrees and radians). They are measured
counter-clockwise starting from the positive x-axis (“upwards”.) Negative angles (not
displayed) are measured clockwise starting from the positive x-axis (“downwards”.)
* For any point 𝑥, 𝑦 on the circle: 𝒙 = cos𝜽 and 𝒚 = sin𝜽 (displayed outside the circle).
For example,
−
1
2𝜋
3
2𝜋
= cos
, and = sin
.
2
3
2
3
* The most important quadrant is in the upper right of the circle (1st quadrant of the
x-y plane). If you know the information given in this part of the circle (angles and their sine
and cosine values) then by symmetry you can determine the information given for the
rest of the circle.
Trigonometry for Calculus: Self Assessment
6. Match each formula to one of the given graphs, below.
(a) 𝑦 = sin𝑥 + 1
(b) 𝑦 = sin2𝑥
(c) 𝑦 = 2sin2𝑥
(e) 𝑦 = cos2𝑥
(f) 𝑦 = 2cos𝑥
(g) 𝑦 = 2cos
(I) (b) 𝒚 = sin𝟐𝒙
!
(d) 𝑦 = sin
!
(h) 𝑦 =
!
!
cos!!!
!
(II) (e) 𝒚 = cos𝟐𝒙
𝒙
(III)
(g) 𝒚 = 𝟐cos
(V)
(a) 𝒚 = sin𝒙 + 𝟏
(VII)
(d) 𝒚 = sin
𝟐
𝒙
𝟐
(IV)
(f) 𝒚 = 𝟐cos𝒙
(VI)
(c) 𝒚 = 𝟐sin𝟐𝒙
(VIII) (h) 𝒚 =
cos𝒙!𝟏
𝟐
Trigonometry for Calculus: Self Assessment
7. Let 𝑓 𝑥 = cos𝑥, 𝑔 𝑥 = 𝑥 ! + 2, and ℎ 𝑥 =
(a) 𝑓(𝑔 𝑥 )
!
(b) 𝑔(𝑓 𝑥 )
(a) 𝑓 𝑔 𝑥
= cos(𝐱 𝟐 + 𝟐)
(b) 𝑔 𝑓 𝑥
= cos 𝟐 𝐱 + 𝟐 (c) ℎ 2 − 𝑓 ! 𝑥
=𝑔
(e) 𝑓 𝑔 ℎ 𝑥
=𝑓
!
cos!
!
!
!
. Find and simplify the given expression.
(c) ℎ(2 − 𝑓 𝑥 )
=
𝟏
cos𝐱
(d) 𝑔(ℎ 𝑓 𝑥 )
(e) 𝑓(𝑔 ℎ 𝑥 )
𝟏
= ℎ 2 − cos ! 𝑥 =
(d) 𝑔 ℎ 𝑓 𝑥
!
2 ! cos 𝟐 𝐱
+𝟐
+ 2 = 𝐜𝐨𝐬
𝟏
𝐱
+𝟐
8. A ladder 10 feet long leans against a vertical wall. Let 𝜃 be the angle between the top of the ladder and
the wall, and let 𝑥 be the distance from the bottom of the ladder to the wall. (Assume that 𝑥 > 0.)
(a) Make a diagram that incorporates the given information.
(b) Express sin𝜃 in terms of 𝑥.
(c) If the ladder slides down the wall, how does 𝑥 change: does it increase, decrease, or stay the same?
How does sin𝜃 change? How does 𝜃 change?
(d) Express cos𝜃 in terms of 𝑥.
(e) If the ladder slides down the wall, how does cos𝜃 change?
(a)
𝜃 10 x 𝒙
opp
(b) In a right triangle, sin𝜽 = hyp. For the triangle in the diagram we thus have sin𝜽 = 𝟏𝟎.
(c) If the ladder slides down the wall (see figures)
𝒙
x increases, so sin𝜽 = 𝟏𝟎 increases and 𝜽 increases.
(d) Using the identity cos ! 𝜃 + sin! 𝜃 = 1, cos ! 𝜃 = 1 −
!!
!""
, so cos𝜽 =
𝟏−
𝒙𝟐
𝟏𝟎𝟎
=
adj
𝟏𝟎𝟎!𝒙𝟐
𝟏𝟎
𝒚
.
(Alternate method: let y be the side of the triangle along the wall. Then cos𝜽 = hyp = 𝟏𝟎. Now use
the Pythagorean Theorem.)
(e) If the ladder slides down the wall, then 𝜃 increases which means
adj
𝒚
cos𝜽 decreases. This is because cos𝜽 =
= , and as 𝜃 increases, y decreases.
hyp
𝟏𝟎