Answers2025-2026 FBISD - Precalculus A CRTrigonometric Double Angle Identities

Trigonometric Double Angle Identities Answers

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1
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Which derivation correctly uses the cosine sum identity to prove the cosine double angle identity?

A
A 2-column table with 3 rows. Column 1 has entries 1, 2, 3. Column 2 is labeled Step with entries cosine (2 x) = cosine (x + x), = cosine (x) cosine (x) minus sine (x) sine (x), = cosine squared (x) minus sine squared (x).
Option A
B
A 2-column table with 3 rows. Column 1 has entries 1, 2, 3. Column 2 is labeled Step with entries cosine (2 x) = cosine (x + x), = cosine (x) cosine (x) + sine (x) sine (x), = cosine squared (x) + sine squared (x).
Option B
C
A 2-column table with 3 rows. Column 1 has entries 1, 2, 3. Column 2 is labeled Step with entries cosine (2 x) = cosine (x + x), = sine (x) sine (x) + cosine (x) cosine (x), = sine squared (x) + cosine squared (x).
Option C
D
A 2-column table with 3 rows. Column 1 has entries 1, 2, 3. Column 2 is labeled Step with entries cosine (2 x) = cosine (x + x), = sine (x) sine (x) minus cosine (x) cosine (x), = sine squared (x) minus cosine squared (x).
Option D
2
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What is the value of 2cos2(75°) – 1?

A
Negative StartFraction StartRoot 3 EndRoot Over 2 EndFraction
Option A
B
Negative one-half
Option B
C
One-half
Option C
D
StartFraction StartRoot 3 EndRoot Over 2 EndFraction
Option D
3

Which expression is equal to ?

Question illustration
A
Negative sine (StartFraction pi Over 20 EndFraction)
Option A
B
Negative sine (StartFraction pi Over 5 EndFraction)
Option B
C
Sine StartFraction pi Over 5 EndFraction
Option C
D
Sine StartFraction pi Over 20 EndFraction
Option D
4

Suppose where . What is the value of tan(2x)?

Question illustration
A
Negative StartFraction 2 StartRoot 8 EndRoot Over 7 EndFraction
Option A
B
Negative StartRoot 8 EndRoot
Option B
C
StartRoot 8 EndRoot
Option C
D
StartFraction 2 StartRoot 8 EndRoot Over 7 EndFraction
Option D
5

Consider the derivation of an alternate form of the cosine double angle identity.

Question illustration
A
In step 1, cos(2x) is equal to cos2(x) + sin2(x).
B
In step 2, sin2(x) should have been replaced with 1 + cos2(x).
C
In step 3, cos2(x) – 1 – cos2(x) should be cos2(x) – 1 + cos2(x).
D
In step 4, 2cos2(x) – 1 should be 1 – 2cos2(x).
6

Consider the derivation of sin(2x).

Question illustration
A
Step 2 should read = cos(x)cos(x) – sin(x)sin(x).
B
Step 2 should read = sin(x)cos(x) – cos(x)sin(x).
C
Step 3 should read = – 2sin(x)cos(x).
D
Step 3 should read = 2sin(x)cos(x).
7

Let sin(2x) – sin(x) = 0, where 0 ≤ x < 2π. What are the possible solutions for x?

A
Left-brace StartFraction pi Over 6 EndFraction, StartFraction 5 pi Over 6 EndFraction Right-brace
Option A
B
Left-brace StartFraction pi Over 3 EndFraction, StartFraction 5 pi Over 3 EndFraction right-brace
Option B
C
Left-brace StartFraction pi over 6 EndFraction, StartFraction pi Over 2 EndFraction, StartFraction 5 pi Over 6 EndFraction, StartFraction 3 pi Over 2 EndFraction right-brace
Option C
D
Left-brace 0, StartFraction pi Over 3 EndFraction, pi, StartFraction 5 pi Over 3 EndFraction right-brace
Option D
8

Which expression is equivalent to cos(4x)?

A
2(2cos2(x) – 1)2 – 1
B
2cos2(x) – 1
C
2cos2(x) – 4
D
4cos2(x)
9

Suppose where 180° < x < 270°. What is the value of sin(2x)?

Question illustration
A
StartFraction 7 Over StartRoot 53 EndRoot EndFraction
Option A
B
StartFraction 14 Over StartRoot 53 EndRoot EndFraction
Option B
C
StartFraction 28 Over 53 EndFraction
Option C
D
StartFraction 49 Over 53 EndFraction
Option D

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