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Math Help - Help on trig identity.

  1. #1
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    Help on trig identity.

    Establish each identity.
    cos(2x)/1 + sin(2x) = cotx - 1/ cotx + 1

    Now in the solutions manual, this is what they have in the middle:
    (cosx- sinx)(cosx+sinx)/cos^2x + sin^2x + 2sinxcosx =
    (cosx- sinx)(cosx+sinx)/(cosx+sinx)(cosx+sinx)

    How did they get rid of 2sinxcosx?
    Thanks.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ktcrew View Post
    Establish each identity.
    cos(2x)/1 + sin(2x) = cotx - 1/ cotx + 1

    Now in the solutions manual, this is what they have in the middle:
    (cosx- sinx)(cosx+sinx)/cos^2x + sin^2x + 2sinxcosx =
    (cosx- sinx)(cosx+sinx)/(cosx+sinx)(cosx+sinx)

    How did they get rid of 2sinxcosx?
    Thanks.
    they factorized. remember, a^2 + 2ab + b^2 = (a + b)(a + b) = (a + b)^2

    if you take for instance, a = \sin x and b = \cos x, then you get \sin^2 x + 2 \sin x \cos x + \cos^2 x = (\sin x + \cos x)(\sin x + \cos x)
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  3. #3
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    Ok thanks. Man, I have trouble seeing the forest through the trees.
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