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Math Help - Prove that secx=[2(cosxsin2x-sinxcos2x)]/sin2x

  1. #1
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    Prove that secx=[2(cosxsin2x-sinxcos2x)]/sin2x




    Prove that secx=[2(cosxsin2x-sinxcos2x)]/sin2x

    explain it to me fully please details please~!
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  2. #2
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    While you are proving this, where are you getting lost?

    Which step are you struggling with?

    If you are struggling to begin, I normally suggest using your double angle formulae so that all of the trig functions have an argument of just x instead of 2x.
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  3. #3
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    I have up to:
    [2sin(2x-x)]/sin2x
    [2sin(2x-x)]/2sinxcosx

    Questions:
    how do I cancel 2sin(2x-x) so that it equals 1/cosx?

    does 2sin(2x-x) really equal 2sin(x) or any number for x for that matter?
    so can I cancel out the 2sin(2x-x)/2sinx?
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  4. #4
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    Quote Originally Posted by skeske1234 View Post
    I have up to:
    [2sin(2x-x)]/sin2x
    [2sin(2x-x)]/2sinxcosx

    Questions:
    how do I cancel 2sin(2x-x) so that it equals 1/cosx?

    does 2sin(2x-x) really equal 2sin(x) or any number for x?
    so can I cancel out the 2sin(2x-x)/2sinx?
    yes ... 2\sin(2x-x) = 2\sin{x}
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