hello-show-that-sin-5pi-18-cos-13pi-18-1-2-sin-4pi-9-

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

hello

show that

\sin \frac{5 π}{18} \cos \frac{13 π}{18} = - \frac{1}{2} \sin \frac{4 π}{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 π}{18} - \frac{13 π}{18}) + \sin (\frac{5 π}{18} + \frac{13 π}{18})) =

= \frac{1}{2} (\sin (- \frac{4 π}{9}) + \sin π) = \frac{1}{2} \sin (- \frac{4 π}{9}) =

= - \frac{1}{2} \sin \frac{4 π}{9}

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