Show that $\sin2nx=\sin((2n+1)x)\cos x-\cos((2n+1)x)\sin x$.[2] Hence prove by mathematical i

Question

Show that $\sin2nx=\sin((2n+1)x)\cos x-\cos((2n+1)x)\sin x$.[2] Hence prove by mathematical induction that $\cos x+\cos3x+\cos5x+\cdots+\cos((2n-1)x)=\frac{\sin2nx}{2\sin x}$ for all $n\in\mathbb Z^+$, $\sin x\ne0$.[7] Find $\sin5^\circ\cos5^\circ+\sin5^\circ\cos15^\circ+\sin5^\circ\cos25^\circ$.[3]

Accepted Answer

(c) $\frac14$.

Trigonometry - Compound and Double-Angle Identities

Trigonometry - Compound and Double-Angle Identities

Solution:

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