Double Angle Formulas Quiz: Solving Equations and Evaluating Values Using Double-Angle Formulas

Starting from cos(2θ) = cos²θ − sin²θ and substituting cos²θ = 1 − sin²θ gives cos(2θ) = (1 − sin²θ) − sin²θ = 1 − 2sin²θ. This form is useful when the problem involves sinθ rather than cosθ. Option D reverses the sign and is incorrect.

The answer is True. sin(2θ) = 2sinθ cosθ. If cosθ = 0, then sin(2θ) = 2sinθ × 0 = 0 regardless of the value of sinθ. This occurs at θ = π/2 + kπ, where cos(2θ) = 1 − 2sin²(π/2 + kπ) = 1 − 2(1) = −1, confirming that 2θ lands at π + 2kπ where sin = 0.

cos(2θ) = 2cos²θ − 1 = 2(−3/5)² − 1 = 2(9/25) − 1 = 18/25 − 1 = −7/25. Note that squaring cosθ removes the negative sign, so the quadrant of θ affects the sign of cosθ but not cos²θ. The result is negative because cos²θ = 9/25 < 1/2.

The three standard forms of cos(2θ) are cos²θ − sin²θ (option A), 1 − 2sin²θ (option B), and 2cos²θ − 1 (option C). Option D, 2sin²θ − 1, is the negative of the correct sine-only form: 2sin²θ − 1 = −(1 − 2sin²θ) = −cos(2θ), not cos(2θ).

The answer is True. tan(2θ) = 2tanθ/(1 − tan²θ). When tanθ = 1, the denominator becomes 1 − 1² = 0, making the expression undefined. Geometrically, when tanθ = 1, θ = π/4 + kπ, and 2θ = π/2 + 2kπ, which are exactly the angles where tangent is undefined.

tan(2θ) = 2tanθ/(1 − tan²θ) = 2(−3/4)/(1 − 9/16) = (−3/2)/(7/16) = (−3/2)(16/7) = −24/7. The negative result is expected because 2θ falls in Quadrant III or the second half of Quadrant II when θ is a small negative angle in Quadrant IV.

cos(2θ) = 1/2 means 2θ = π/3 + 2kπ or 2θ = −π/3 + 2kπ, giving θ = π/6 + kπ and θ = −π/6 + kπ. On [0, 2π): from π/6 + kπ: π/6 and 7π/6. From −π/6 + kπ: 5π/6 and 11π/6. Verification: cos(π/3) = 1/2 ✓, cos(5π/3) = 1/2 ✓, cos(7π/3) = cos(π/3) = 1/2 ✓, cos(11π/3) = cos(5π/3) = 1/2 ✓.

The answer is True. This is the standard double-angle formula for tangent, derived from tan(α+β) = (tanα + tanβ)/(1 − tanα tanβ) with α = β = θ. The formula is defined whenever 1 − tan²θ ≠ 0, i.e. when tanθ ≠ ±1.

The answer is False. The correct cosine-only double-angle formula is cos(2θ) = 2cos²θ − 1, not 2cos²θ + 1. A quick check confirms the error: at θ = 0, cos(0) = 1 but 2cos²(0) + 1 = 2(1) + 1 = 3 ≠ 1. The sign between the two terms must be minus, not plus.

cos(2θ) = −1 means 2θ = π + 2kπ, so θ = π/2 + kπ. On [0, 2π) this gives θ = π/2 and θ = 3π/2. Verification: cos(2 × π/2) = cos(π) = −1 ✓ and cos(2 × 3π/2) = cos(3π) = −1 ✓. Options B (θ=0) and C (θ=π) give cos(0) = 1 and cos(2π) = 1, not −1.

The double-angle formula for sine is sin(2θ) = 2sinθ cosθ, derived from the sum formula sin(α+β) = sinα cosβ + cosα sinβ with α = β = θ. Option A gives the Pythagorean identity, not a double-angle formula. Option C omits the factor of 2. Option D uses squares rather than the product of sine and cosine.

cos(2θ) = 0 means 2θ = π/2 + kπ, so θ = π/4 + kπ/2. On [0, 2π) this gives θ = π/4, 3π/4, 5π/4, and 7π/4 — four solutions in total. There are four rather than two because 2θ covers the interval [0, 4π), allowing two full cycles of cosine to reach zero.

In Quadrant IV, cosθ > 0, so cosθ = 12/13. Then tanθ = sinθ/cosθ = (−5/13)/(12/13) = −5/12. Using the double-angle formula: tan(2θ) = 2tanθ/(1 − tan²θ) = 2(−5/12)/(1 − 25/144) = (−10/12)/(119/144) = (−5/6)(144/119) = −120/119.

The correct double-angle formula is tan(2θ) = 2tanθ / (1 − tan²θ), not 1 + tan²θ. Option B is valid since tan = sin/cos applied to 2θ. Option C is valid since it uses sin(2θ) = 2sinθ cosθ and cos(2θ) = cos²θ − sin²θ. Option D has the wrong sign in the denominator — 1 + tan²θ = sec²θ, which gives a different and incorrect expression.

Let u = 2θ, so u ∈ [−2π, 2π]. sin(u) = √2/2 gives u = π/4 + 2kπ or u = 3π/4 + 2kπ. For k=0: u = π/4 → θ = π/8 and u = 3π/4 → θ = 3π/8. For k=−1: u = −7π/4 → θ = −7π/8 and u = −5π/4 → θ = −5π/8. Verification: sin(2 × π/8) = sin(π/4) = √2/2 ✓; sin(2 × 3π/8) = sin(3π/4) = √2/2 ✓; sin(2 × (−7π/8)) = sin(−7π/4) = √2/2 ✓; sin(2 × (−5π/8)) = sin(−5π/4) = √2/2 ✓. Option A is incorrect — 5π/8 and −3π/8 produce sin values of −√2/2, not √2/2.

2 × 30° = 60°. cos(60°) = 1/2. Alternatively, using the double-angle formula: cos(2 × 30°) = 1 − 2sin²(30°) = 1 − 2(1/4) = 1 − 1/2 = 1/2. Both methods confirm the answer.

Using the cosine-only form: cos(2θ) = 2cos²θ − 1 = 2(2/3)² − 1 = 2(4/9) − 1 = 8/9 − 1 = −1/9. Even though θ is in Quadrant I, cos(2θ) can be negative when cos²θ < 1/2, which occurs when cosθ < 1/√2 ≈ 0.707. Since 2/3 ≈ 0.667 < 0.707, the result is negative.

Using the sine-only form: cos(2θ) = 1 − 2sin²θ = 1 − 2(√3/2)² = 1 − 2(3/4) = 1 − 3/2 = −1/2. The quadrant of θ determines sin²θ and confirms the calculation; sin²θ = 3/4 regardless of sign.

sin(2θ) = 2sinθ cosθ = 2 × (3/5) × (4/5) = 2 × 12/25 = 24/25. Since θ is in Quadrant I, 2θ is in Quadrant I or II and sin(2θ) is positive, confirming the positive result.

sin(2θ) = 0 means 2θ = kπ for integer k, so θ = kπ/2. On [0, 2π) the distinct solutions are θ = 0, π/2, π, and 3π/2. There are four solutions because 2θ ranges over [0, 4π), doubling the number of solutions compared to sinθ = 0.