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Math Help - Proving confuction Identity

  1. #1
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    Proving confuction Identity

    Hello,

    Can someone please help me with this problem?

    Problem:

    csc(pi/2 - u) = sec u

    I understand that csc = 1/sin

    how do I factor ----- 1/sin(pi/2-u)

    That's where I'm basically stuck.

    Let me know, thanks!
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  2. #2
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    Quote Originally Posted by l flipboi l View Post
    Hello,

    Can someone please help me with this problem?

    Problem:

    csc(pi/2 - u) = sec u

    I understand that csc = 1/sin

    how do I factor ----- 1/sin(pi/2-u)

    That's where I'm basically stuck.

    Let me know, thanks!
    csc \left(\frac{\pi}{2} - u \right) = \frac{1}{sin\left(\frac{\pi}{2} - u \right)}

    sin(A+B) = sinAcosB + cosAsinB

    For  \theta = \frac{\pi}{2}

    sin \theta = 1 and cos \theta = 0
    Last edited by e^(i*pi); October 31st 2009 at 02:30 PM. Reason: typo
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  3. #3
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    Quote Originally Posted by l flipboi l View Post
    Hello,

    Can someone please help me with this problem?

    Problem:

    csc(pi/2 - u) = sec u

    I understand that csc = 1/sin

    how do I factor ----- 1/sin(pi/2-u)

    That's where I'm basically stuck.

    Let me know, thanks!
    \sin\left(\frac{\pi}{2} - u\right) = \sin\left(\frac{\pi}{2}\right)\cos(u) - \cos\left(\frac{\pi}{2}\right)\sin(u)

    continue ...
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  4. #4
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    So the law of distributing fractions is:

    a/b (c+d) = a / (bc+bd) ?
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  5. #5
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    Quote Originally Posted by l flipboi l View Post
    So the law of distributing fractions is:

    a/b (c+d) = a / (bc+bd) ?
    don't complicate things ...

    just evaluate/simplify the expression

    \sin\left(\frac{\pi}{2} - u\right) = \sin\left(\frac{\pi}{2}\right)\cos(u) - \cos\left(\frac{\pi}{2}\right)\sin(u)<br />

    and finish the proof.
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