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Math Help - [Quick Question]Rewriting Pythagorean Identity

  1. #1
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    Exclamation [Quick Question]Rewriting Pythagorean Identity



    I need help with this problem. Please show your work and explain how you rewrote the Pythagorean equation into what they are asking. Thank you so much.
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  2. #2
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    Quote Originally Posted by florx View Post


    I need help with this problem. Please show your work and explain how you rewrote the Pythagorean equation into what they are asking. Thank you so much.

    Remember \sin ^2t+\cos^2t = 1

     <br />
2\cos^2t-1 = 2\cos^2t-(\sin ^2t+\cos^2t)= 2\cos^2t-\sin ^2t-\cos^2t = \cos^2t-\sin ^2t<br />
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  3. #3
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    Thank you so much for your quick answer for a). I had forgot you could make 1 = sin t + cos t

    Anyways can you check to see if I did b) correctly?

    b) cos 2t = (cos t)^2 - (sin t)^2
    cos 2t = -sin^2t + cos^2t
    cos^2t -sin^2t
    (1-sin^2t) + (cos^2t -1)
    cos 2t = -sin^2t + cos^2t
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  4. #4
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    Quote Originally Posted by florx View Post
    Thank you so much for your quick answer for a). I had forgot you could make 1 = sin t + cos t

    Anyways can you check to see if I did b) correctly?

    b) cos 2t = (cos t)^2 - (sin t)^2
    cos 2t = -sin^2t + cos^2t
    cos^2t -sin^2t
    (1-sin^2t) + (cos^2t -1)
    cos 2t = -sin^2t + cos^2t
    If you look at pickslides' post, you can see that he has already done part(b) for you.

    2\cos^2t-1 = 2\cos^2t-(\sin ^2t+\cos^2t)= 2\cos^2t-\sin ^2t-\cos^2t = \cos^2t-\sin ^2t

    Do you know that 2\cos^2t-1 = \cos2t? In fact, this is the part(a) of your question!

    And your part (b) is thus proved!
    Last edited by harish21; April 12th 2010 at 07:50 PM.
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    Oh thanks I didn't notice that.
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