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

1. freckles

sounds like you are trying to show this is an identity

2. anonymous

Which I am trying to figure out now lol

3. anonymous

it looks like i should be using 2 sin theta and cos theta

4. freckles

is the above written correctly?

5. freckles

seems like there could be a cube or something missing

6. anonymous

yeah it suppose to be cosine cubed which i have there

7. freckles

well see if you can continue from here $8 \sin(x) \cos^3(x)-4 \sin(x)\cos(x) \\ \text{ I would start by factoring out } 4 \sin(x)\cos(x) \\ 4 \sin(x) \cos(x)[\cos^2(x)-1] \\ 2 \cdot 2 \sin(x)\cos(x)[-1(1-\cos^2(x))]$

8. freckles

like right here I see a couple of identities you can use

9. freckles

oops missed a 2

10. freckles

$8 \sin(x) \cos^3(x)-4 \sin(x)\cos(x) \\ \text{ I would start by factoring out } 4 \sin(x)\cos(x) \\ 4 \sin(x) \cos(x)[2\cos^2(x)-1] \\ 2 \cdot 2 \sin(x)\cos(x)[-1(1-2 \cos^2(x))]$

11. freckles

beautiful :)

12. 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

13. 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

14. anonymous

This is what I did

15. anonymous

so is this right?

16. 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

17. freckles

or on the first line

18. 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

19. freckles

you just need some ( ) in there

20. freckles

this is exactly what I would do to fix your ( ) issue: $4 \sin(x)\cos(x)(2\cos^2(x)-1) \text{ for first line; just drop that one \between the } \\ 4 \text{ and } \sin(x) \\ \\$

21. freckles

the second line ... I would change that first ( to a * and then also put ( ) around the 2cos^2(x)-1 like so: $2 \cdot 2 \sin(x)\cos(x)(2 \cos^2(x)-1)$

22. anonymous

Thanks soo much again!! I swear so much work under one problem lol @freckles