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Math Help - Proving

  1. #1
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    Proving

    I like proving but this one, I got stuck because of time pressure. It was the last question and I was hurrying since I only got 5 mins left. I didn't get to finish my solution so I may have gotten a few points off it.

    here it goes:

    1/cos^2B = [cosB sin2B - sinB cos2B]/cosB sinB

    so what I did first was:

    1/cos^2B = cos(B+2B)/cosB sinB


    then I got stuck so I repeated it then came out with:

    1/cos^2B = sinB - cosB/cosB sinB by subtracting the functions (idk if that's allowed) then I tried to break that down. I am still stuck.

    anyone who can help me on how the sol'n goes??
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  2. #2
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    Hello pathurr19
    Quote Originally Posted by pathurr19 View Post
    I like proving but this one, I got stuck because of time pressure. It was the last question and I was hurrying since I only got 5 mins left. I didn't get to finish my solution so I may have gotten a few points off it.

    here it goes:

    1/cos^2B = [cosB sin2B - sinB cos2B]/cosB sinB

    so what I did first was:

    1/cos^2B = cos(B+2B)/cosB sinB


    then I got stuck so I repeated it then came out with:

    1/cos^2B = sinB - cosB/cosB sinB by subtracting the functions (idk if that's allowed) then I tried to break that down. I am still stuck.

    anyone who can help me on how the sol'n goes??
    I get a slightly different result.

    \frac{\cos B\sin2B-\sin B\cos2B}{\cos B \sin B}=\frac{\sin(2B-B)}{\cos B \sin B}

    =\frac{\sin B}{\sin B\cos B }

    =\frac{1}{\cos B }

    Grandad
    Last edited by Grandad; October 21st 2009 at 11:58 PM. Reason: Simpler method
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  3. #3
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    Sep 2006
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    Just like Grandad, I got different results.

    \frac{1}{cos^{2B}} = \frac{cosB sin2B - sinB cos2B}{cosB sinB}

    \frac{1}{cos^{2B}} = \frac{cosB (2sinB cosB) - sinB (2cos^{2}B - 1)}{cosB sinB}

    \frac{1}{cos^{2B}} = \frac{(2sinB cos^{2}B) - sinB (2cos^{2}B - 1)}{cosB sinB}

    \frac{1}{cos^{2B}} = \frac{(2sinB cos^{2}B) - (2sinBcos^{2}B - sin B)}{cosB sinB}

    \frac{1}{cos^{2B}} = \frac{sin B}{cosB sinB}

    \frac{1}{cos^{2B}} = \frac{sin B}{cosB sinB}

    \frac{1}{cos^{2B}} = \frac{1}{cosB}
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