Question

cos3x/sinx+sin3x/cosx=cotx-tanx prove

Answer 1

I would use the identities for the cos of a sum and sin of a sum, with 1x and 2x being the quantities summed. Then break the cot and tan functions down into the basic definitions in terms of sin and cos, and you should be able to get to the end.

Answer 2

well, i can't touch both sides i must only change in one side. the problem is i get stuck when i use double angel formula.

Answer 3

No, you can touch both sides, as long as you do the same thing to each of them.

Answer 4

\[\frac{\sin (3 x)}{\cos (x)}=\frac{\sin (x) \left(2 \cos ^2(x)-1\right)+2 \sin (x) \cos (x) \cos (x)}{\cos (x)}\]\[\frac{\cos (3 x)}{\sin (x)}=\left(1-2 \sin ^2(x)\right) \cos (x)-2 \sin (x) \sin (x) \cos (x)\]\[\cot (x)-\tan (x)=\frac{\cos (x)}{\sin (x)}-\frac{\sin (x)}{\cos (x)}\]

I think you should be able to work it out from that.

Answer 5

Sorry, lost my denominator on the second line, it should have been:
\[\frac{\cos (3 x)}{\sin (x)}=\frac{\left(1-2 \sin ^2(x)\right) \cos (x)-2 \sin (x) \sin (x) \cos (x)}{\sin (x)}\]

Answer 6

Don't actually need to touch the right hand side, though it is perfectly legal to do so.