Double-Angle Formulas: Identity Proof & Derivation Quiz

  • 11th Grade
Reviewed by Cierra Henderson
Cierra Henderson, MBA |
K-12 Expert
Review Board Member
Cierra is an educational consultant and curriculum developer who has worked with students in K-12 for a variety of subjects including English and Math as well as test prep. She specializes in one-on-one support for students especially those with learning differences. She holds an MBA from the University of Massachusetts Amherst and a certificate in educational consulting from UC Irvine.
 , MBA
By Thames
T
Thames
Community Contributor
Quizzes Created: 8156 | Total Attempts: 9,586,195
| Attempts: 11 | Questions: 20 | Updated: Jan 22, 2026
Please wait...
Question 1 / 21
🏆 Rank #-- ▾
Score 0/100
1) Which identity is correct for all real θ?

Explanation

Step 1: Start from the sum identity: sin(α+β)=sinα cosβ+cosα sinβ.

Step 2: Set α=β=θ → sin(2θ)=sinθ cosθ+cosθ sinθ.

Step 3: Combine like terms → sin(2θ)=2 sinθ cosθ.

So, the final answer is sin(2θ)=2 sinθ cosθ.

Submit
Please wait...
About This Quiz
Double-angle Formulas: Identity Proof & Derivation Quiz - Quiz

Get ready to explore how one angle can double the excitement! In this quiz, you’ll prove and derive the double-angle formulas for sine, cosine, and tangent — including sin(2θ) = 2sinθcosθ and cos(2θ) = cos²θ − sin²θ. You’ll also see how these identities connect to the Pythagorean relationship sin²θ +... see morecos²θ = 1. Through calculation and reasoning, you’ll practice transforming and simplifying trigonometric expressions step by step.

see less

2)
You may optionally provide this to label your report, leaderboard, or certificate.
2) Which is a correct form of the double-angle identity for cosine?

Explanation

Step 1: Start with cos(2θ)=cos²θ−sin²θ.

Step 2: Substitute sin²θ=1−cos²θ → cos(2θ)=cos²θ−(1−cos²θ).

Step 3: Simplify → cos(2θ)=2 cos²θ−1.

So, the final answer is cos(2θ)=2 cos²θ−1.

Submit
3) Simplify 1−2 sin²θ.

Explanation

Step 1: Recall cos(2θ)=1−2 sin²θ.

Step 2: The given matches the identity exactly.

So, the final answer is cos(2θ).

Submit
4) Evaluate sin(2θ) if sinθ = 3/5 and θ is in Quadrant II.

Explanation

Step 1: Use sin²θ+cos²θ=1 → cos²θ=1−(3/5)²=1−9/25=16/25.

Step 2: cosθ=±4/5; in Quadrant II, cosθ is negative → cosθ=−4/5.

Step 3: Apply sin(2θ)=2 sinθ cosθ = 2×(3/5)×(−4/5)=−24/25.

So, the final answer is −24/25.

Submit
5) If cosθ = −4/5 and θ is in Quadrant II, find cos(2θ).

Explanation

Step 1: Use cos(2θ)=2 cos²θ−1.

Step 2: cos²θ=(−4/5)²=16/25.

Step 3: Substitute → cos(2θ)=2(16/25)−1=32/25−25/25=7/25.

So, the final answer is 7/25.

Submit
6) Derive tan(2θ) from tan(α+β) formula.

Explanation

Step 1: Set α=β=θ → tan(2θ)=(tanθ+tanθ)/(1−tan²θ).

Step 2: Simplify numerator → 2 tanθ.

So, the final answer is tan(2θ)=(2 tanθ)/(1−tan²θ).

Submit
7) If cos(2θ)=2 cos²θ−1, find cos²θ.

Explanation

Step 1: cos(2θ)=2 cos²θ−1.

Step 2: Add 1 → cos(2θ)+1=2 cos²θ.

Step 3: Divide by 2 → cos²θ=(1+cos(2θ))/2.

So, the final answer is (1+cos(2θ))/2.

Submit
8) If tanθ=1/2 and θ is acute, find tan(2θ).

Explanation

Step 1: tan(2θ)=(2 tanθ)/(1−tan²θ).

Step 2: Substitute tanθ=1/2 → (2×1/2)/(1−1/4)=1/(3/4)=4/3.

So, the final answer is 4/3.

Submit
9) Express sin²θ in terms of cos(2θ).

Explanation

Step 1: From cos(2θ)=1−2 sin²θ, rearrange → 2 sin²θ=1−cos(2θ).

Step 2: Divide by 2 → sin²θ=(1−cos(2θ))/2.

So, the final answer is (1−cos(2θ))/2.

Submit
10) Given sinθ=−5/13 and θ in Quadrant IV, find cos(2θ).

Explanation

Step 1: cos²θ=1−sin²θ=1−(25/169)=144/169 → cosθ=12/13 (Q IV → positive).

Step 2: cos(2θ)=cos²θ−sin²θ=(144/169)−(25/169)=119/169.

So, the final answer is 119/169.

Submit
11) Simplify sin(2θ) + 2 sinθ cosθ

Explanation

Step 1: Recall sin(2θ)=2 sinθ cosθ.

Step 2: Substitute → sin(2θ)+2 sinθ cosθ = (2 sinθ cosθ)+2 sinθ cosθ = 4 sinθ cosθ.

Step 3: Write 4 sinθ cosθ as 2 (2 sinθ cosθ) = 2 sin(2θ).

So, the final answer is 2 sin(2θ).

Submit
12) Which identity is equivalent to cos(2θ)?

Explanation

Step 1: Start from cos(2θ)=cos²θ−sin²θ.

Step 2: Substitute sin²θ=1−cos²θ.

Step 3: Simplify → cos(2θ)=cos²θ−(1−cos²θ)=2 cos²θ−1.

So, the final answer is 2 cos²θ − 1.

Submit
13) If 0 < θ < π/2 and cos(2θ)=1/5, find sinθ.

Explanation

Step 1: Use cos(2θ)=1−2 sin²θ.

Step 2: Substitute 1/5 = 1 − 2 sin²θ.

Step 3: Simplify → 2 sin²θ = 4/5 → sin²θ = 2/5.

Step 4: Since θ is acute, sinθ > 0 → sinθ = √(2/5).

So, the final answer is √(2/5).

Submit
14) Which identity is true for all real θ?

Explanation

Step 1: Recall the three standard forms of cos(2θ):



cos(2θ)=cos²θ−sin²θ, = 1−2 sin²θ, = 2 cos²θ−1.

Step 2: Among these, 1−2 sin²θ matches option B.

So, the final answer is cos(2θ)=1−2 sin²θ.


Submit
15) If cosθ=12/13 and θ is in Quadrant IV, find sin(2θ).

Explanation

Step 1: Find sinθ using sin²θ = 1 − cos²θ → sinθ = −√(1−(12/13)²)=−5/13 (Q IV → negative).

Step 2: Use sin(2θ)=2 sinθ cosθ = 2 × (−5/13) × (12/13)=−120/169.

So, the final answer is −120/169.

Submit
16) Starting from cos(2θ)=cos²θ−sin²θ and using sin²θ+cos²θ=1, derive an alternate form.

Explanation

Step 1: Replace sin²θ with 1−cos²θ.

Step 2: Simplify → cos(2θ)=cos²θ−(1−cos²θ)=2 cos²θ−1.

So, the final answer is cos(2θ)=2 cos²θ−1.

Submit
17) Given cotθ = x and x > 0, which expression equals csc²θ?

Explanation

Step 1: Use the Pythagorean identity 1 + cot²θ = csc²θ.

Step 2: Substitute cotθ = x → csc²θ = 1 + x².

So, the final answer is 1 + x².

Submit
18) Find cos(2θ) if θ = 15°.

Explanation

Step 1: Compute 2θ = 30°.

Step 2: cos(30°)=√3/2.

So, the final answer is √3/2.

Submit
19) Which expression equals sin(2θ) in terms of x=cos (θ)?

Explanation

Step 1: Recall the double-angle identity for sine:

sin(2θ) = 2sin(θ)cos(θ).

Step 2: Since x = cos(θ), substitute cos(θ) = x.

Step 3: Use the Pythagorean identity sin²(θ) + cos²(θ) = 1 to get sin(θ) = √(1 − x²).

Step 4: Substitute both into the formula:

sin(2θ) = 2(√(1 − x²))(x).

So, the final answer is 2x√(1 − x²).

Submit
20) Over [0, 2π], how does y = sin(2θ) compare to y = sinθ?

Explanation

Step 1: The period of sinθ is 2π.

Step 2: For sin(2θ), doubling the angle halves the period.

Step 3: Therefore, the new period is π.

So, the final answer is that sin(2θ) has a period π.

Submit
×
Saved
Thank you for your feedback!
View My Results
Cierra Henderson |MBA |
K-12 Expert
Cierra is an educational consultant and curriculum developer who has worked with students in K-12 for a variety of subjects including English and Math as well as test prep. She specializes in one-on-one support for students especially those with learning differences. She holds an MBA from the University of Massachusetts Amherst and a certificate in educational consulting from UC Irvine.
Cancel
  • All
    All (20)
  • Unanswered
    Unanswered ()
  • Answered
    Answered ()
Which identity is correct for all real θ?
Which is a correct form of the double-angle identity for cosine?
Simplify 1−2 sin²θ.
Evaluate sin(2θ) if sinθ = 3/5 and θ is in Quadrant...
If cosθ = −4/5 and θ is in Quadrant II, find...
Derive tan(2θ) from tan(α+β) formula.
If cos(2θ)=2 cos²θ−1, find cos²θ.
If tanθ=1/2 and θ is acute, find tan(2θ).
Express sin²θ in terms of cos(2θ).
Given sinθ=−5/13 and θ in Quadrant IV, find...
Simplify sin(2θ) + 2 sinθ cosθ
Which identity is equivalent to cos(2θ)?
If 0 < θ < π/2 and cos(2θ)=1/5, find sinθ.
Which identity is true for all real θ?
If cosθ=12/13 and θ is in Quadrant IV, find sin(2θ).
Starting from cos(2θ)=cos²θ−sin²θ...
Given cotθ = x and x > 0, which expression equals...
Find cos(2θ) if θ = 15°.
Which expression equals sin(2θ) in terms of x=cos (θ)?
Over [0, 2π], how does y = sin(2θ) compare to y = sinθ?
play-Mute sad happy unanswered_answer up-hover down-hover success oval cancel Check box square blue
Alert!