SOLUTION: Hope you can help me: Solve sin2y = cos4y for y, where 0 <= y < 360

Algebra ->  Trigonometry-basics -> SOLUTION: Hope you can help me: Solve sin2y = cos4y for y, where 0 <= y < 360      Log On


   



Question 62148: Hope you can help me:
Solve sin2y = cos4y for y, where 0 <= y < 360

Answer by joyofmath(189) About Me  (Show Source):
You can put this solution on YOUR website!
Solve sin2y = cos4y for y, where 0 <= y < 360
Let x = 2y. Then, sinx+=+cos2x.
There's a trig identity: cos2x+=+1-2%28sinx%29%5E2
We use that identity and get: sin%28x%29+=+1-2%28sinx%29%5E2
Let some variable, say g, = sinx.
Then, g+=+1-2g%5E2.
Or, 2g%5E2%2Bg+=+1
Or, 2g%5E2%2Bg-1+=+0.
This factors into %282g-1%29%28g%2B1%29+=+0.
So, g=1%2F2 or g=-1.
So, since g+=+sinx, sinx+=+1%2F2 or sinx+=+-1.
Since x = 2y, sin%282y%29+=+1%2F2 or sin%282y%29+=+-1.
So, 2y = 270 degress or 2y = 30 degrees.
So, the answer is y = 135 degrees or y = 15 degrees.
Let's verify the answer:
sin(2*135) = sin(270) = -1 and cos(4*135) = cos(540) = cos(180) = -1.
sin(2*15) = sin(30) = 1/2 and cos(60) = 1/2.