## anonymous one year ago Are (sin3x/sin x cos x)=4 cos x-sec x and (sin 3x-sin x)/(cos 3x+cos x)= tan x identities?

1. sohailiftikhar

use the formulas to find this

2. sohailiftikhar

sin3x=3sinx-4sin^3

3. sohailiftikhar

ok u can put 3sinx-4sin^3x rather than sin3x (3sinx-4sin^3x)/(sinx)(cosx) now take sinx common from above equation sinx(3-4sin^2x)/sinx cosx =(3-4sin^2x)/cosx

4. sohailiftikhar

now you can put sin^2x=1-cos^2x (3-4(1-cos^2x)/cosx (3-4+4cos^2x)/cosx -1+4cos^2x/cosx -1(1-4cos^2x)/cosx

5. sohailiftikhar

as u know cos2x=1-2cos^2x and cos4x=1-4cos^2x now put 1-4cos^2x as cos4x

6. sohailiftikhar

-cos4x/cosx

7. sohailiftikhar

so part a is incorrect

8. anonymous

Thank you for pointing it out step by step! I have a really hard time figuring out which identities to use and how to simplify them. I hope you don't mind me asking, what step was secant changed?

9. sohailiftikhar

are u asking about second equation ? and no problem it's my pleasure to help u .

10. anonymous

Yes please, I'd love for help with the second equation!

11. sohailiftikhar

ok lemme check

12. sohailiftikhar

ok sin3x=3sinx-4cos^3x cos3x=4cos^3x-3cosx

13. sohailiftikhar

we will solve both separately first then we'll multiply them ok u can put 3sinx-4sin^3x rather than sin3x (3sinx-4sin^3x-sinx) now take sinx common from above equation sinx(3-4sin^2x-1) =sinx(2-4sin^2x)

14. sohailiftikhar

and for cos3x+cosx as cos3x=4cos^3x-3cosx so 4cos^3x-3cosx+cosx now take cosx common cosx(4cos^2x-3+1)=cosx(4cos^2x-2)

15. sohailiftikhar

now combine both equations sinx(2-4sin^2x)/cosx(4cosx^2-2)

16. sohailiftikhar

take 2 from above and 2 from below 2(sinx)(1-2sin^2x)/2(cosx)(2cos^2-1) =sinx/cosx=tanx and yes it is true

17. anonymous

Thank you! The second one was true but apparently the first one was also true, not false. I appreciate the time it took to go through all the steps, though! It makes understanding the concept a lot easier.

18. sohailiftikhar

may be first one can true .. perhaps i did some mistake in solving this

19. anonymous

It might be the secant part in (sin3x/sin x cos x)=4 cos x-sec x ? I'm not very good at trig identities, though....

20. sohailiftikhar

ok try this urself by using the formulas which i told u . u can do this now just don't skip any step ok

21. anonymous

$3sinx-4\sin^3/\sin x \cos x =4 \cos x-\sec x$ $(3-4\sin^2x)/cosx=4 \cos x - \sec x$ $(3-4(1-\cos^2x)/cosx=4\cos x-\sec x$ $-1+4\cos^2x/cosx$ $-1-4\cos^2x/\cos=4\cos x-\sec x$ $4\cos^2x-1/\cos=4\cos^2x-1/\cos$ I tried my best c': it's just my best is sub par....