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Thread: Proving Trig Identities

  1. #1
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    Proving Trig Identities

    Hey Guys and Girls!
    I need some help answering one of my homework questions. This is from Grade 12u Advanced Functions.
    I've done other examples, but this one seems to confuse me.

    Prove the identity:

    sinxcosxtanx = 1 - cos^2x

    This is what I've done so far:
    Left Side:
    =sinxcosx(sinx/cosx)

    and from here I don't know what to do, but my guess would be:
    = sin^2x / cos^2x

    Which means both sides are not equal?

    Right Side:
    1 - cos^2x
    =sin^2x

    Thanks for any help you can give me!
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  2. #2
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    You've done it right but you've just made a small error. I'll right it out in full for you:
    Left side
    $\displaystyle \sin x \cos x \tan x = \sin x \cos x (\frac{\sin x}{\cos x}) $
    The above way of setting it out might allow you to see your error. Both $\displaystyle \cos x $ should cancel each other out

    $\displaystyle = \sin x * \sin x $
    $\displaystyle = \ sin^2 x $
    $\displaystyle = 1 - \cos^2 x $
    $\displaystyle = right side $
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  3. #3
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    Yes that is what I was thinking, but how come the cosx's cancel out yet the sinx's become sin^2x?
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  4. #4
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    You already have $\displaystyle \sin x \cos x $ and $\displaystyle \tan x = \frac{\sin x}{\cos x} $ so multiplying the two it becomes:

    $\displaystyle \sin x \cos x * \frac{\sin x}{\cos x} $

    $\displaystyle \frac{\sin x * \cos x * \sin x}{\cos x} $

    leaving:

    $\displaystyle \sin^2 x $

    Just like using numbers:

    $\displaystyle \frac{4}{5} * \frac{3}{4} $

    $\displaystyle \frac{4*3}{5*4} $

    here the 4 cancels leaving:

    $\displaystyle \frac{3}{5} $

    Does that make it clearer?
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  5. #5
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    Oh I get it....Thank you so much
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