Question

Prove The Following Trig Identity:

cos(4x)+cos(2x)
---------------- = cot(3x)
sin(4x)+sin(2x)

so:

cos^2(2x)-sin^2(2x)+cos^2(x)-sin^2(x)
----------------------------------
2sin(2x)cos(2x)+2sin(x)cos(x)


cos^2(3x)-sin^2(3x)
-------------------------
2sin(2x)cos(2x)+2sin(x)cos(x)

Answer 1

I did something wrong after this, and ended up with cot(6x) :<

Answer 2

I know it's true by definition, I just don't know how to prove it otherwise. ..Maybe I'm going at it the wrong way, I don't know.

Answer 3

Alright, the first step is to use the Sum to product rule for both the top and the bottom.

Answer 4

\[\frac{ 2\cos \left( \frac{ 4x+2x }{ 2 } \right)\cos \left( \frac{ 4x-2x }{ 2 } \right) }{ 2\sin \left( \frac{ 4x+2x }{ 2 } \right)\cos \left( \frac{ 4x-2x }{ 2 } \right) }\]

Answer 5

That will simplify into\[\frac{ 2\cos(3x)\cos(x) }{ 2\sin(3x)\cos(x) }\]
So we can cancel out our like terms 2, and cos(x) and that leaves us with\[\frac{ \cos(3x) }{ \sin(3x) } \rightarrow \cot(3x)\]

Answer 6

Ok, thank you.

Answer 7

You're welcome.