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Math Help - Trig prob

  1. #1
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    Question solved: Trig prob

    4 cosec22x + 4 cosec22x cot22x, need to simplify this and the answer is suppose 2 be 4cosec 4 2x
    Last edited by Sundae; April 27th 2009 at 09:42 AM. Reason: {Solved it}
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  2. #2
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    Quote Originally Posted by Sundae View Post
    4 cosec22x + 4 cosec22x cot22x, need to simplify this and the answer is suppose 2 be 4cosec 4 2x
    Hi

    \frac{4}{\sin^2 2x} + \frac{4}{\sin^2 2x} \:\frac{\cos^2 2x}{\sin^2 2x} = \frac{4}{\sin^2 2x}\:\left(1 + \frac{\cos^2 2x}{\sin^2 2x}\right) = \frac{4}{\sin^2 2x}\:\frac{\sin^2 2x + \cos^2 2x}{\sin^2 2x}

    \frac{4}{\sin^2 2x} + \frac{4}{\sin^2 2x} \:\frac{\cos^2 2x}{\sin^2 2x} = \frac{4}{\sin^4 2x}
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  3. #3
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    Hello, Sundae!

    Did you really expect us to be able to read that??


    4 cosec22x + 4 cosec22x cot22x
    need to simplify this and the answer is suppose 2 be 4cosec 4 2x

    When u r riting math things, donut use a "2" 4 da word "to"
    . . or use a "4" 4 da word "for".
    Math iz hard enuf 2 read widout cutesy abbreviations n spellings 2 interpret.


    Factor: . 4\csc\!22x\left(1 + \cot\!22x\right) \;=4\csc\!22x\,\csc\!22x \;=\;4\csc\!42x . . . Got it?





    Okay, I'm kidding . . . I will take a wild guess that 22x is actually ^22x


    We have: . 4\csc^2\!2x + 4\csc^2\!2x\cot^2\!2x

    \text{Factor: }\;4\csc^2\!2x\underbrace{\left[1 + \cot^2\!2x\right]}_{\text{This is }\csc^2\!2x}

    . . . . = \;4\csc^2\!2x\cdot\csc^2\!2x \;=\;4\csc^4\!2x

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