Double-Angle Formulas: Identity Proof & Derivation Quiz

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Quizzes Created: 7116 | Total Attempts: 9,522,086
| Questions: 20 | Updated: Oct 31, 2025
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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.

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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

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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.

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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).

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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.

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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.

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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).

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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.

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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.

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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.

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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.

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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).

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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.

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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).

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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.


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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.

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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.

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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.

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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.

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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).

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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.

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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?
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