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Math Help - csc (A - B)

  1. #1
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    csc (A - B)

    If A = 45 degrees and B = arcsin(3/5), find the exact value of csc (A - B).
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  2. #2
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    Quote Originally Posted by magentarita View Post
    If A = 45 degrees and B = arcsin(3/5), find the exact value of csc (A - B).
    Hints:

    \csc{X} = \frac{1}{\sin{X}}


    \sin{(\alpha - \beta)} = \sin{\alpha}\cos{\beta}- \cos{\alpha}\sin{\beta}.
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  3. #3
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    Yes but...

    Quote Originally Posted by Prove It View Post
    Hints:

    \csc{X} = \frac{1}{\sin{X}}


    \sin{(\alpha - \beta)} = \sin{\alpha}\cos{\beta}- \cos{\alpha}\sin{\beta}.
    My problem with this question is arcsin(3/5). In the given formula, A = 45 and B = arcsin(3/5). I can't go on without knowing the value of arcsin(3/5).
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  4. #4
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    Quote Originally Posted by magentarita View Post
    My problem with this question is arcsin(3/5). In the given formula, A = 45 and B = arcsin(3/5). I can't go on without knowing the value of arcsin(3/5).
    Yes you can.

    Remember that \sin{(\arcsin{\theta})} = \theta and \cos{\theta} = \sin{(90^\circ - \theta)}.
    Last edited by Chris L T521; May 6th 2009 at 06:56 PM. Reason: fixed LaTeX
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  5. #5
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    ok

    I will play with this question and get back to you later.
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  6. #6
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    Quote Originally Posted by magentarita View Post
    I will play with this question and get back to you later.
    Note that B = \arcsin \left( \frac{3}{5}\right) \Rightarrow \sin B = \frac{3}{5}.

    And \sin B = \frac{3}{5} \Rightarrow \cos B = \frac{4}{5} (the postive value is taken because the range of the arcsin function is angles in the 4th and 1st quadrants and cos is postive in these two quadrants).
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