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Question Number 75147 by mathocean1 last updated on 07/Dec/19

$$\mathrm{hello}$$$$\mathrm{show}\mathrm{that}$$$$\sin \frac{5\pi }{18}\cos \frac{13\pi }{18}=-\frac{1}{2}\sin \frac{4\pi }{9}$$

Answered by MJS last updated on 07/Dec/19

$$\sin a\cos b=\frac{1}{2}(\sin (a-b)+\sin (a+b))$$$$\frac{1}{2}(\sin (\frac{5\pi }{18}-\frac{13\pi }{18})+\sin (\frac{5\pi }{18}+\frac{13\pi }{18}))=$$$$=\frac{1}{2}(\sin (-\frac{4\pi }{9})+\sin \pi )=\frac{1}{2}\sin (-\frac{4\pi }{9})=$$$$=-\frac{1}{2}\sin \frac{4\pi }{9}$$

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