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Math Help - Trig identities help

  1. #1
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    Trig identities help

    Guys, i swear im going to fail functions this year T_T

    I've been paying attention in class, but i dont get what the teacher is saying at all.

    Were supposed to prove that these 2 things are equal.
    Can someone pleeease show me how to do this?
    because ive tried mixing them around but its not working at all.

    so heres the question i have to prove

    1 - (sin^2 y)/(1+(1/tan y)) - cos^2 y/1+tan y = sin y cos y

    the teacher also wrote

    recall: tan y = sin y/cos y

    sin^2 y + cos ^2 y = 1

    id really appreciate some help T_T thanks so much guys
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  2. #2
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    Quote Originally Posted by gazoline View Post
    1 - (sin^2 y)/(1+(1/tan y)) - cos^2 y/1+tan y = sin y cos y
    I am going to assume that you meant

    1-\frac{\sin^2y}{1+1/\tan y}-\frac{\cos^2y}{1+\tan y}=\sin y\cos y

    (with the last \tan y included in the denominator).

    Begin by combining the fractions on the left side.

    1-\frac{\sin^2y}{1+1/\tan y}-\frac{\cos^2y}{1+\tan y}

    =\frac{1+\tan y}{1+\tan y}-\frac{\tan y\sin^2y}{\tan y\,(1+1/\tan y)}-\frac{\cos^2y}{1+\tan y}

    =\frac{1+\tan y}{1+\tan y}-\frac{\tan y\sin^2y}{1+\tan y}-\frac{\cos^2y}{1+\tan y}

    =\frac{1+\tan y-\tan y\sin^2y-\cos^2y}{1+\tan y}

    Factor,

    =\frac{1+\tan y\left(1-\sin^2y\right)-\cos^2y}{1+\tan y}

    =\frac{1+\tan y\cos^2y-\cos^2y}{1+\tan y}

    Convert to sines and cosines.

    =\frac{1+\sin y\cos y-\cos^2y}{1+(\sin y/\cos y)}

    =\frac{\sin^2y+\sin y\cos y}{(\sin y+\cos y)/\cos y}

    Now just simplify and reduce the fraction, and you're there.
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  3. #3
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    thx so much for the quick response i really aprpeciate it
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