Download Trig Identitiy Review Answer Key

Honors Advanced Math
Trig Identity Review Packet
Name: ANSWER KEY
For the Trig Identitity test you should be able to:
• Derive any of the angle sum, double angle, half angle or power reducing identities
• Solve equations by using the identities to simplify the equations
• Prove identities (other than the basics)
• Use the identities to solve problems
The practice test online includes a reference sheet with several identities. You will not be given
a sheet like this for the real exam. You must know the basic identities and be able to derive the
others from them.
1. Use the cofunction identities and odd-even identities to prove that sin(π − x) = sin(x) .
⎛ π ⎛
⎛ ⎛ π
⎛
⎞⎞
⎛ π
⎞
π ⎞⎞
π ⎞
sin(π − x) = sin⎜ − ⎜ x − ⎟⎟ = cos⎜ x − ⎟ = cos⎜ −⎜ − x ⎟⎟ = cos⎜ − x ⎟ = sin(x)
⎝
⎠⎠
⎝ 2
⎠
2 ⎠⎠
2 ⎠
⎝ 2 ⎝
⎝ ⎝ 2
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2. Use the cofunction identities and odd-even identities to prove that cos(π − x) = −cos(x) .
⎛ π ⎛
⎛ ⎛ π
⎛
⎞⎞
⎛ π
⎞
π ⎞⎞
π ⎞
cos(π − x) = cos⎜ − ⎜ x − ⎟⎟ = sin⎜ x − ⎟ = sin⎜ −⎜ − x ⎟⎟ = −sin⎜ − x ⎟ = −cos(x)
⎝
⎠⎠
⎝ 2
⎠
2 ⎠⎠
2 ⎠
⎝ 2 ⎝
⎝ ⎝ 2
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3. Use the identity in problem 1 to prove that in any triangle ABC, sin(A + B) = sin(C).
In a triangle, A + B + C = π so C = π – (A + B) so sin(C) = sin(π – (A + B)) = sin(A + B)
4. Use the identities in problems 1 and 2 to find an identity for simplifying tan(π – x).
sin(π − x )
sin(x)
tan(π − x ) =
=
= −tan(x)
cos(π − x ) −cos(x)
5. A rhombus is a quadrilateral with equal sides. The diagonals of a
rhombus bisect the angles of the rhombus and are perpendicular
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bisectors of each other. Let ∠ALF = α and ∠LHA = β Each side is
of length s.
⎛ α ⎞ HL
⎛ α ⎞ FA
a) Show that cos⎜ ⎟ =
and sin⎜ ⎟ =
⎝€
2 ⎠ 2s
€⎝ 2 ⎠ 2s
F
H
HL ⋅ FA
b) Show that sin(α ) =
(Start with the area of ΔLAF,
2s2
€
calculated
from the trig€triangle area and from the geometric triangle area)
L
A
c) Show that the area of the rhombus is s2 sin(2β ) (∠AHF = 2β and the area of the rhombus is
€
the area of the two congruent triangles.)
d) Use the fact that the diagonals are perpendicular to show that its area is 2s2 sin(β ) ⋅ cos(β )
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Area HALF = 2*(0.5 * (FA * 0.5 HL)) and use the results from part a
e) Use your answers from (c) and (d) to obtain a formula for sin(2β ) set the areas equal and solve.
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Review Packet for Trig Identities
page 2
6. In ΔABC, the measure of ∠B is twice the measure of ∠C.
a. Use the law of sines to show that b = 2c cos C
c
b
b
Start with
and solve for c.
=
=
sinC sin B 2sinC cosC
b. Use the law of cosines to show that b2 = c(a + c)
Start with c2 = a2 + b2 – 2abcosC and replace cosC using the result in part a. Factor out a b2/c
€ the last two terms, subtract a2 from c2 and factor the difference of squares. You should
from
now see where to go to finish.
7. Solve each of the following for 0 ≤ x < 2π (except e and f) using trigonometric identities:
π 5π
π 5π 7π 11π
1
a. sin x cos x =
b. cos2x = 5sin 2 x − cos 2 x x = , , ,
x= ,
4 4
6 6 6 6
2
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c. tan2x + tan x = 0
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π 2π 4 π 5π
d. sin2x ⋅ sec x + 2cos x = 0
x = 0, π , , , ,
3 3 €
3 3
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e. cos−1 2x = sin−1 x
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x=
5
5
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f.
x=
3π 7π
,
4 4
3
tan−1 2x = sin−1 x x = 0, ±
€ 2
values.
€ 8. Here are letter name definitions for some trigonometric
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A = sin(0.1 π)
Β = cos(0.1 π)
C = sin(0.2 π)
D = cos(0.23 π)
E = cos(0.27 π)
F = tan(0.3 π)
Express each of the following trigonometric values in terms of A, B, C, D, E, and/or F. For
example, if csc(0.1 π) were given as a problem, the answer would be csc(0.1 π) = 1A . You may
not use trigonometric inverses in your answers.
a. cot(0.1 π) = B / A
b. cos(0.3 π) = cos(0.5 π – 0.2 π) = sin(0.2 π) = C
c. cos(0.73 π) = cos(π – 0.27 π) = –cos(0.27 π) = –E
d. sin(–0.3 π) = –sin(0.3 π) = –tan(0.3 π) ∗ cos(0.3 π) = –F * C
e. tan(1.4 π) = tan(π + 0.4 π) = tan(0.4 π) = tan (0.5 π – 0.1 π) = cot(0.1 π) = B / A
f. cos(0.5 sin-1(cos(0.3 π))) = cos(0.5 sin-1(C)) = cos(0.5 * 0.2 π) = cos(0.1 π) = B