Math 1080-001 - Midterm 3 Review - Practice Problems
Ch. 5: p. 369-370 (Self-Test) #1, 2, 4-10all
p. 370-371 (Practice for Calculus) #17-26all, 37-43all, 60
Ch. 6: p. 421 (Self-Test) #1-9all
p. 422-423 (Practice for Calculus) #1-4all, 9-16all, 21, 22, 24, 29, 31, 36, 37, 38, 41, 43, 53, 56, 57, 58, 59
1. Trig. Basics and Trig. Functions
Find the exact value of each expression.
1-1. Find the reference angle: θ = 5π
6
1-2. Convert to degrees: θ = − 11π
3
1-3. Covert to radians: θ = 240◦
1-4. tan π
1-5. sec( 3π
4 )
1-6. sin( 5π
3 )
1-7. cos π3 + 4 sin 2π
3
1-8. 3 cos2 150◦ − 2
Find the exact value of all six trig. functions.
1-9. Given cot θ = 12
5 and cos θ < 0.
1-10. Given sin θ = − 72 and tan θ < 0.
2. Graphs of Trig. Functions
Idenify the period, amplitude, domain, range, and phase shift. Then graph (Your graph should include at
least 1 full period).
2-1. y = 2 sin(4x)
2-2. y = tan(x − π3 )
2-3. y = 5 cos(2x − π2 )
Identify the period, amplitude, domain, and range. Then write the equation for each graph.
2-4.
2-5.
2-6.
1
3. Inverse Trig. Functions
True or False. Explain your reasoning.
3-A. The domain of y = sin−1 x is − π2 ≤ x ≤
3-B. cos(sin−1 0) = 1 and sin(cos−1 0) = 1.
π
2
Find the exact value.
3-1. sin−1 (1)
3-2. cos−1 (− 12 )
3-3. cos(arccos 7π
6 )
3-4. tan−1 (tan 4π
3 )
3-5. sin−1 (sin π6 )
3-6. tan(sin−1 (− 45 ))
3-7. cos(sin−1 ( 35 ) − cos−1 ( 12 ))
3-8. sin(2 tan−1 ( 43 ))
4. Solve Triangles (Law of Sines/Law of Cosines) and Area of Triangles
True or False. Explain your reasoning.
4-A. An oblique triangle in which two sides and an angle are given always results in at least one triangle.
4-B. Given three sides of a triangle, there is a formula for finding its area.
4-C. In a right triangle, if two sides are know, we can solve the triangle.
4-D. The ambiguous case refers to the fact that when two sides and the angle opposite one of them is known,
sometime the Law of Sines cannot be used.
Solve each triangle.
4-1. α = 30◦ , b=8, and c=10
4-2. a=10, b=6, c=14
4-3. α = 150◦ , β = 15◦ , a=4
4-4. β = 105◦ , γ = 75◦ , c=12
B
C
A
Find the area of each triangle.
4-5. α = 150◦ , β = 15◦ , a=4
4-6. α = 30◦ , b=8, and c=10
2
5. Solve Trig. Equations
True or False. Explain your reasoning.
5-A. Most trigonometric equations have unique solutions.
Solve.
5-1. cos θ = 12
5-2. sin(2θ) = −1
5-3. 2 sin2 θ − 3 sin θ + 1 = 0
5-4. sin θ − sin(2θ) = 0 for − π2 ≤ θ ≤ π
5-5. sin2 θ − cos2 θ = 0 for 0 ≤ θ ≤ 2π
6. Use Trig. Identities
Find the exact value of each expression.
6-1. sin 165◦
6-2. tan 105◦
6-3. cos 5π
12
6-4. sin(− π8 )
6-5. cos 80◦ cos 20◦ + sin 80◦ sin 20◦
150◦
6-6. sin
cos 15◦
6-7. cos2 75◦ + sin2 240◦
7. Prove Trig. Identities
True or False. Explain your reasoning.
7-A. sin(−θ) + sin(θ) = 0 for all θ
7-B. sin(α + β) = sin α + sin β + 2 sin α sin β
7-C. cos 2θ has √
three equivalent forms: sin2 θ − cos2 θ, 1 − 2 sin2 θ, and 2 cos2 θ − 1.
α
7-D. cos α2 = ± 1+cos
, where the + or - sign depends on the angle α2
2
Prove the following identities.
7-1. sin θ csc θ − sin2 θ = cos2 θ
θ
cos3 θ
7-2. 1−sin
sec θ = 1+sin θ
cos(α+β)
7-3. cos
α sin β = cot β − tan α
7-4. sin θ tan θ2 = 1 − cos θ
θ−sin θ cos 3θ
7-5. sin 3θ cossin
=1
2θ
2
1−2 sin θ
7-6. sin θ cos θ = cot θ − tan θ
3
Additional Identity Practice: Prove the following identities.
1. tanθ = cotθtan2 θ
2. tan2 θ − sin2 θ = tan2 θsin2 θ
3.
1−sec2 β
sec2 β
= −sin2 β
4. 1 − sin(2α) =
5.
1+cos(2λ)sec(2λ)
tan(2λ)+sec(2λ)
1−sin2 (2α)
1+sin(2α)
2cos(2λ)
sin(2λ)+1
=
6. (sinθ + cosθ)2 + (sinθ − cosθ)2 = 2
7.
1+cot2 γ
1+tan2 γ
= cot2 γ
8. 1 + sin2 y = 2 − cos2 y
9. sin2 β + tan2 β = sec2 β − cos2 β
10. (cotθ + cscθ)2 =
2
= ( 1−cosx
sinx )
11.
1−cosx
1+cosx
12.
1+tan3 θ
1+tanθ
13.
2tanα+2tanαsecα
tanα+secα+1
14.
tanθ
cotθ
−
secθ+1
secθ−1
= sec2 θ − tanθ
cotθ
tanθ
= tanα + secα − 1
= sec2 θ − csc2 θ
15. (secx + tanx)2 =
cscx+1
cscx−1
4