## anonymous 4 years ago sin8x=8sinxcosxcos2xcos4x proving trig identity?

1. anonymous

$\sin8x=8sinxcosxcos2xcos4x$

2. anonymous

i have LS: sin8x = sin(4x+4x) = 2sin4xcos4x then i have no idea.. should i just keep on expanding? or is there another way of solving this?

3. TuringTest

repeat the process: sin4x=2sin2xcos2x sin2x=2sinxcosx

4. anonymous

just keep doing that? :/ would that even get simplified?

5. anonymous

$\sin 8x = 2 \sin 4x \cos 4x = 4\sin 2x \cos 2x \cos 4x = 8 \sin x \cos x \cos 2x \cos 4x$

6. TuringTest

not sure if you want to call it 'simplified', but it is what you were looking for

7. anonymous

oh. it is! ahaha thanks.

8. anonymous

In general, $\cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A \cdots \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin A }$ $\implies 2^n \sin A \cdot \cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A \cdots \cos 2^{n-1}A= \sin 2^n A$

9. TuringTest

nice^

10. anonymous

Thanks :)