A question about trigonometric identities

One obvious point is that if x= 0, the left side is 1 but the right side does not exist. And if you don't like "does not exist", try evaluating both sides at x= 0.01.

 

Try cross multiplying to get:

[tex]\cos 2x (\sin 3x + \sin x) = \sin 2x (\cos 3x + \cos x)[/tex]

and then apply the half sum/difference identities to all of the sin cos products on each side of the equation. It comes out fairly easy with this step.

Use:
[tex]\sin a \cos b = \frac{1}{2} \left[ \sin(a+b) + \sin(a-b) \right][/tex]

 

An alternate approach is to convert to a polynomial equation in u = e^ix.
For example: cot(2x) = i(u²+u^-2)/(u²-u^-2) = i(u⁴+1)/(u⁴-1).
Do the similar steps for the right hand side, clear the denominators and end up with an identity.

 

You need to end up with sinx and cosx only, using identities for cos2x, etc. I think my method (using representation in e^ix, etc.) might be easier.

 

No, apply the half sum-difference formula (that I gave above) to each of the sin-cos products.