sin(4x)=8 sin(x)cos^3(x)-4 sin(x)cos(x) Proofs are beyond confusing...

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sin(4x)=8 sin(x)cos^3(x)-4 sin(x)cos(x) Proofs are beyond confusing...

Mathematics
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sounds like you are trying to show this is an identity
Which I am trying to figure out now lol
it looks like i should be using 2 sin theta and cos theta

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is the above written correctly?
seems like there could be a cube or something missing
yeah it suppose to be cosine cubed which i have there
well see if you can continue from here \[8 \sin(x) \cos^3(x)-4 \sin(x)\cos(x) \\ \text{ I would start by factoring out } 4 \sin(x)\cos(x) \\ 4 \sin(x) \cos(x)[\cos^2(x)-1] \\ 2 \cdot 2 \sin(x)\cos(x)[-1(1-\cos^2(x))] \]
like right here I see a couple of identities you can use
oops missed a 2
\[8 \sin(x) \cos^3(x)-4 \sin(x)\cos(x) \\ \text{ I would start by factoring out } 4 \sin(x)\cos(x) \\ 4 \sin(x) \cos(x)[2\cos^2(x)-1] \\ 2 \cdot 2 \sin(x)\cos(x)[-1(1-2 \cos^2(x))] \]
beautiful :)
for example guess what 2sin(x)cos(x) or guess what 2cos^2(x)-1 or if you like -1(1-2cos^2(x)) equals
http://www.eeweb.com/tools/math-sheets/images/trigonometry-laws-and-identities-small.png look under double angle identities if you do not recall
This is what I did
so is this right?
I like the last 5 lines and the first 2 lines looks like you factored as I did above just didn't end the ( parenthesis thing on the second line
or on the first line
\[8 \sin(x) \cos^3(x)-4 \sin(x)\cos(x) \\ \text{ I would start by factoring out } 4 \sin(x)\cos(x) \\ 4 \sin(x) \cos(x)[2\cos^2(x)-1] \\ \] but yeah that is essentially what I have above
you just need some ( ) in there
this is exactly what I would do to fix your ( ) issue: \[4 \sin(x)\cos(x)(2\cos^2(x)-1) \text{ for first line; just drop that one \between the } \\ 4 \text{ and } \sin(x) \\ \\\]
the second line ... I would change that first ( to a * and then also put ( ) around the 2cos^2(x)-1 like so: \[2 \cdot 2 \sin(x)\cos(x)(2 \cos^2(x)-1)\]
Thanks soo much again!! I swear so much work under one problem lol @freckles

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