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Having trouble getting this trigonometry problem started

  1. Oct 5, 2006 #1
    The problem is

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

    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.
    Last edited: Oct 5, 2006
  2. jcsd
  3. Oct 5, 2006 #2


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    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 equasion from here. (EDIT - This can be factorised for a solution)
  4. Oct 6, 2006 #3
    Thanks alot acm, I appreciate it. :wink:

    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.
    Last edited: Oct 6, 2006
  5. Oct 6, 2006 #4


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  6. Oct 6, 2006 #5
    Ahhh thats right, I glanced over that handy identity, but should have put more effort into remembering it. Thanks!
  7. Oct 6, 2006 #6
    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 :)
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