Having trouble getting this trigonometry problem started

[Checkfate]

The problem is

Solve sin(2y)=cos(4y) for y, where \\0\leq y \leq 360

This one is tricky, I want to try to equate either sin to cos or cos to sin so I can work with only one trig function, but how? I don't see how to start this problem. I know that the period of sin2y is 720deg or 4pi, and the period of cos4y is 8pi or 1440deg.

 

[acm]

I'd start it by expanding Cos4y to (Cos2y)^2 - (Sin2y)^2.
Sin2y = (Cos2y)^2 - (Sin2y)^2
Sin2y= 1 - (Sin2y)^2 - (Sin2y)^2
2(Sin2y)^2 + Sin2y - 1 = 0

I'd probably use the Binomial theorem to solve the equation from here. (EDIT - This can be factorised for a solution)

 

[Checkfate]

Thanks a lot acm, I appreciate it.

Could someone tell me the rule though? Cos4y definitely does expand out to (Cos2y)^2 - (Sin2y)^2, but how?

I would guess that Cos9y = (Cos3y)^2-(Sin3y)^2, but it doesn't...

I just want something that I can write down in my notes so that I can use it in the future. Thanks. Or if someone could refer me to a website that would be perfect.

 

[acm]

The theorem is Cos2z = (Cosz)^2 - (Sinz)^2.
This therem can be prooven using complex numbers or geometry.
A link with a multitude of trigonometric formulas is: http://en.wikipedia.org/wiki/Double-angle_formula

 

[Checkfate]

Ahhh that's right, I glanced over that handy identity, but should have put more effort into remembering it. Thanks!

 

[Checkfate]

Thats actually neater then it looks :) I played around with it for a bit and realized that Cos(4x) = Cos(2x+2x) = cos(2x)cos(2x)-sin(2x)sin(2x)=(cos(2x))^2-(sin(2x))^2 :)