Question 1

Which is the correct double angle identity for $\sin(2\alpha)$?

Accepted Answer

2sin(alpha)cos(alpha) — Setting $\beta = \alpha$ in $\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$ gives $\sin(2\alpha) = 2\sin\alpha\cos\alpha$.

Question 2

Which is the compound angle identity for $\cos(\alpha - \beta)$?

Accepted Answer

cos(alpha)cos(beta) + sin(alpha)sin(beta) — $\cos(\alpha - \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$ (the sign flips relative to $\cos(\alpha+\beta)$).

Question 3

Which expression equals $\cos(2\alpha)$ using only $\cos$?

Accepted Answer

2cos^2(alpha) - 1 — $\cos(2\alpha) = \cos^2\alpha - \sin^2\alpha = \cos^2\alpha - (1-\cos^2\alpha) = 2\cos^2\alpha - 1$.

Trigonometry

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