Question

sin8x=8sinxcosxcos2xcos4x
proving trig identity?

Answer 1

\[\sin8x=8sinxcosxcos2xcos4x\]

Answer 2

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?

Answer 3

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

Answer 4

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

Answer 5

\[ \sin 8x = 2 \sin 4x \cos 4x = 4\sin 2x \cos 2x \cos 4x = 8 \sin x \cos x \cos 2x \cos 4x \]

Answer 6

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

Answer 7

oh. it is!
ahaha thanks.

Answer 8

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

Answer 9

nice^

Answer 10

Thanks :)