anonymous
  • anonymous
sin(4x)=8 sin(x)cos^3(x)-4 sin(x)cos(x) Proofs are beyond confusing...
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
freckles
  • freckles
sounds like you are trying to show this is an identity
anonymous
  • anonymous
Which I am trying to figure out now lol
anonymous
  • anonymous
it looks like i should be using 2 sin theta and cos theta

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.