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Math Help - Addition Formulas - Question

  1. #1
    Newbie Cdoddsy's Avatar
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    Question Addition Formulas - Question

    Ok, I have the following problem in my practice homework:
    If sin(a) = (-5/13) and tan(a) > 0, find the exact value of sin(a - (pi/6))

    In the back of the book, the following solution is given:
    Since sin(a) = (-15/13) < 0 and tan(a) > 0, a is in QIII, and Cos(a) = -sqr( 1 - (-5/13)^2 ) = (-12/13)

    My question is, where did this come from, "-sqr( 1 - (-5/13)^2 ) = (-12/13)"? What formula or identity are are they using on this step? It sort of reminds me of the formula of a circle, but I am just not sure...and the book does not say.
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  2. #2
    Senior Member
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    Hi Cdoddsy

    It uses :

    \cos ^2 x=1-\sin^2 x
    Last edited by songoku; November 22nd 2009 at 09:11 PM. Reason: typo
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  3. #3
    Newbie Cdoddsy's Avatar
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    Ohh, I get it, because if:

    cos^2(x) = 1 + sin^2(x), then
    cos(x) = sqr(1+sin^2(x))

    Thanks!
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  4. #4
    Senior Member
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    Oops sorry I made mistake.

    It should be :

    \cos ^2 x = 1-\sin^2 x
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