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anonymous
 one year ago
sin(4x)=8 sin(x)cos^3(x)4 sin(x)cos(x)
Proofs are beyond confusing...
anonymous
 one year ago
sin(4x)=8 sin(x)cos^3(x)4 sin(x)cos(x) Proofs are beyond confusing...

This Question is Closed

freckles
 one year ago
Best ResponseYou've already chosen the best response.1sounds like you are trying to show this is an identity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Which I am trying to figure out now lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it looks like i should be using 2 sin theta and cos theta

freckles
 one year ago
Best ResponseYou've already chosen the best response.1is the above written correctly?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1seems like there could be a cube or something missing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah it suppose to be cosine cubed which i have there

freckles
 one year ago
Best ResponseYou've already chosen the best response.1well see if you can continue from here \[8 \sin(x) \cos^3(x)4 \sin(x)\cos(x) \\ \text{ I would start by factoring out } 4 \sin(x)\cos(x) \\ 4 \sin(x) \cos(x)[\cos^2(x)1] \\ 2 \cdot 2 \sin(x)\cos(x)[1(1\cos^2(x))] \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1like right here I see a couple of identities you can use

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[8 \sin(x) \cos^3(x)4 \sin(x)\cos(x) \\ \text{ I would start by factoring out } 4 \sin(x)\cos(x) \\ 4 \sin(x) \cos(x)[2\cos^2(x)1] \\ 2 \cdot 2 \sin(x)\cos(x)[1(12 \cos^2(x))] \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1for example guess what 2sin(x)cos(x) or guess what 2cos^2(x)1 or if you like 1(12cos^2(x)) equals

freckles
 one year ago
Best ResponseYou've already chosen the best response.1http://www.eeweb.com/tools/mathsheets/images/trigonometrylawsandidentitiessmall.png look under double angle identities if you do not recall

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[8 \sin(x) \cos^3(x)4 \sin(x)\cos(x) \\ \text{ I would start by factoring out } 4 \sin(x)\cos(x) \\ 4 \sin(x) \cos(x)[2\cos^2(x)1] \\ \] but yeah that is essentially what I have above

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you just need some ( ) in there

freckles
 one year ago
Best ResponseYou've already chosen the best response.1this is exactly what I would do to fix your ( ) issue: \[4 \sin(x)\cos(x)(2\cos^2(x)1) \text{ for first line; just drop that one \between the } \\ 4 \text{ and } \sin(x) \\ \\\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1the second line ... I would change that first ( to a * and then also put ( ) around the 2cos^2(x)1 like so: \[2 \cdot 2 \sin(x)\cos(x)(2 \cos^2(x)1)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks soo much again!! I swear so much work under one problem lol @freckles
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