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Math Help - Verifying Trigonometric Identity

  1. #1
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    Verifying Trigonometric Identity

    Hello.

    I'm currently doing chapter reviews for my final exam, and I've come across a trigonometric identity that I have seen before, but don't recall how to verfiy. If you can give me any help and explain it, it will be greatly appreciated. The identity is below.

    \sqrt{\frac{1-sin\theta}{1+sin\theta}}=\frac{1-sin\theta}{|cos\theta|}

    Thank you for your time.

    Bryan
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    Hint : \sqrt{\frac{1-\sin\theta}{1+\sin\theta}}=\sqrt{\frac{(1-\sin\theta)(1-\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}}
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  3. #3
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    Hint : \sqrt{\frac{1-\sin\theta}{1+\sin\theta}}=\sqrt{\frac{(1-\sin\theta)(1-\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}}
    Yeah, my gut was saying to multiply the radical by its conjugate, but I'm really not sure how to get rid of the radical itself. Not to mention I dont know where |\cos\theta| comes from.
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  4. #4
    Moo
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    Hello,

    Quote Originally Posted by mathgeek777 View Post
    Yeah, my gut was saying to multiply the radical by its conjugate, but I'm really not sure how to get rid of the radical itself. Not to mention I dont know where |\cos\theta| comes from.
    1-\sin^2x=\cos^2x (this is the second hint for the denominator )
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  5. #5
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    <br />
\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}=\sqrt{\frac{(1-\sin\theta)(1-\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}}<br />
=\sqrt{\frac{(1-\sin\theta)(1-\sin\theta)}{1-\sin^2}}
    <br />
=\sqrt{\frac{(1-\sin\theta)(1-\sin\theta)}{\cos^2 \theta}}=\frac{\sqrt{(1-\sin\theta)(1-\sin\theta)}}{\sqrt{\cos^2\theta}}<br />
=\frac{1-\sin\theta}{|\cos^2\theta|}<br />
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