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Thread: how to start/ simplify these trigonometric functions

  1. #1
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    how to start/ simplify these trigonometric functions

    csc A - sin A to cos A cot A

    and

    sin A cot^2 A (sec - 1) to sin A
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  2. #2
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    Quote Originally Posted by xahjika View Post
    csc A - sin A to cos A cot A

    and

    sin A cot^2 A (sec - 1) to sin A
    $\displaystyle \csc(A) - \sin(A) = \frac{1}{\sin(A)} - \sin(A) $

    $\displaystyle = \frac{1}{\sin(A)} - \frac{\sin^2(A)}{\sin(A)} $

    $\displaystyle = \frac{\cos^2{A}}{\sin(A)} = \frac{\cos(A) \times \cos(A)}{\sin(A)} $

    $\displaystyle = \frac{\cos(A)}{\sin(A)} \times \cos(A) $
    Last edited by Mush; Jan 22nd 2009 at 07:44 AM.
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  3. #3
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    Are you sure you've written the second one out properly?
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  4. #4
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    Quote Originally Posted by Mush View Post
    Are you sure you've written the second one out properly?
    oops sorry

    sin (A) cot^2 (A) (sec^2 (A) - 1) to sin (A)
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  5. #5
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    Quote Originally Posted by xahjika View Post
    oops sorry

    sin (A) cot^2 (A) (sec^2 (A) - 1) to sin (A)
    Consider the identity:

    $\displaystyle \sin^2(A) + \cos^2(A) = 1 $

    Divide it by $\displaystyle \cos^2(A) $

    $\displaystyle \tan^2(A) + 1 = \sec^2(A) $

    $\displaystyle \sec^2(A) - 1 = \tan^2(A) $

    Hence your problem becomes:

    $\displaystyle \sin(A) \cot^2(A) \tan^2(A) $

    And you know that $\displaystyle \tan^2(A) = \frac{1}{\cot^2(A)}$

    Does that help?
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  6. #6
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    yes it is.. thanks a lot... coz signs bothers me
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