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Math Help - Trig equation

  1. #1
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    Trig equation

    Can anyone help here

    If sin4x/sin2y + cos 4x/cos2y =1,
    prove sin4y/sin2x + cos4y/cos2x =1

    All the 2 and 4 are powers of sin and cos
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  2. #2
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    Quote Originally Posted by math123x View Post
    Can anyone help here

    If sin4x/sin2y + cos 4x/cos2y =1,
    prove sin4y/sin2x + cos4y/cos2x =1

    All the 2 and 4 are powers of sin and cos

    Sin^4y/Sin^2x + Cos^4y/Cos^2x = 1

    First, cross multiply the denominators so that you have the LCD giving you:

    (Sin^4y)(Cos^2x)/(Sin^2x)(Cos^2x) + (Cos^4y)(Sin^2x)/(Sin^2x)(Cos^2x)

    Combine those because they have the same denominator and you get:

    ((Sin^4y)(Cos^2x) + (Cos^4y)(Sin^2x))/((Sin^2x)(Cos^2x))

    Now you can break about Sin^4y and Cos^4y so that they are (Sin^2y)(Sin^2y) and (Cos^2y)(Cos^2y)

    Now you have:
    <br />
((Sin^2y)(Sin^2y)(Cos^2x) + (Cos^2y)(Cos^2y)(Sin^2x))/((Sin^2x)(Cos^2x))

    Because Sin^2 \Theta + Cos^2 \Theta = 1 you can combine all the Sin^2y + Cos^2y into 1 and eliminate them leaving you with:

    (Cos^2x + Sin^2x)/((Sin^2x)(Cos^2x))

    From here you can break it apart into:

    (Cos^2x/((Sin^2x)(Cos^2x))) + (Sin^2x/((Sin^2x)(Cos^2x)))

    Cos^2x from the first half cancels with the denominator leaving you with 1/Sin^2x and the same with Sin^2x in the second half leaving you with 1/Cos^2x

    Now you have left:

    ((1/Cos^2x) + (1/Sin^2x)) = 1

    Cross multiply again to get the LCD, then combine to get:

    Sin^2x/((Cos^2x)(Sin^2x)) + Cos^2x/((Cos^2x)(Sin^2x))

    Because Sin^2 \Theta  + Cos^2 \Theta  = 1 you can reduce both denominators to 1 which gives you:

     1/1 + 1/1 = 1

    There you are! If you have any questions feel free to ask.

    Ah crap, on moment, need to recheck my answer, it comes up as two. sorry.

    Well, I've thought about it for a while and come up with a complete block. If someone else could point me in another direction I would be much obliged. Sorry.

    The only way I can think of doing it, is if you take:

    Sin^2x/((Cos^2x)(Sin^2x)) + Cos^2x/((Cos^2x)(Sin^2x))

    then cancel Sin^2x on the left, and Cos^2x on the right and combine the denominator getting:

    1/(Cos^2x + Sin^2x) = 1

    and because Sin^2 \Theta  + Cos^2 \Theta  = 1 you get 1/1

    The problem is, I don't think you can combine denominators like that which I tried to do before anyway.
    Last edited by Mezzlegasm; February 20th 2009 at 05:26 PM.
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  3. #3
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    wait, what? xD

    I tried to solve it yesterday.. ended up with 2 sheets of paper all written over. :P
    the explanation given above has some serious holes in it. (too tired to point them out at the moment)
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  4. #4
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    Quote Originally Posted by metlx View Post
    wait, what? xD

    I tried to solve it yesterday.. ended up with 2 sheets of paper all written over. :P
    the explanation given above has some serious holes in it. (too tired to point them out at the moment)
    Yeah, I'm in the same situation.

    It has a lot of holes because I was doing it on the computer instead of paper, typing it out, which sucks.

    I didn't realize what I did wrong about about two hours later, and I can't think of another way to do it. I think it's impossible, haha.

    Also I did it like 3-4 months ago, so my mind wasn't on topic immediately.
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