1. ## Prove The Identity

sin8x= (8sinx)(cosx)(cos2x)(cos4x)

i have been trying the double angle formula on sin8x and not having any luck any help is appreciated!

2. Originally Posted by superhume
sin8x= (8sinx)(cosx)(cos2x)(cos4x)

i have been trying the double angle formula on sin8x and not having any luck any help is appreciated!
sin(8x) = sin2(4x)

you should know that sin2x = 2sinxcosx. Applying this to the above, you have

sin2(4x) = 2sin(4x).cos(4x)

again,

sin(4x) = sin2(2x) = 2 sin(2x)cos(2x), so you have,

sin(8x) = 2 [ 2 sin(2x)cos(2x)] cos(4x) = 4 sin(2x)cos(2x)cos(4x)

again put sin2x = 2sinxcosx in the above, and you have your answer

3. Hello, superhume!

Prove: . $\sin8x \:=\: 8\sin x\cos x\cos2x\cos4x$
We can work it from the right side . . .

$\text{We have: }\:8\sin x\cos x\cos2x\cos4x \;=\;4\underbrace{(2\sin x\cos x)}_{\text{This is }\sin2x}\cos2x\cos4x$

. . . . . . . . . . . . . . . . . . . . . $=\;4\sin2x\cos2x\cos4x$

. . . . . . . . . . . . . . . . . . . . . $=\;2\underbrace{(2\sin2x\cos2x)}_{\text{This is }\sin4x}\cos4x$

. . . . . . . . . . . . . . . . . . . . . $=\;\underbrace{2\sin4x\cos4x}_{\text{This is }\sin8x}$

. . . . . . . . . . . . . . . . . . . . . $=\;\sin8x$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We can work it from the left side . . .

$\text{We have: }\;\sin8x \;=\;2\underbrace{\sin4x}\cos4x$

. . . . . . . . $=\; 2\overbrace{(2\sin2x\cos2x)}\cos4x$

. . . . . . . . $=\;4\underbrace{\sin2x}\cos2x\cos4x$

. . . . . . $=\;4\overbrace{(2\sin x\cos x)}\cos2x\cos4x$

. . . . . . $=\;8\sin x\cos x\cos2x\cos4x$

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# 4sin2xcos2x is equal to what?

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