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Math Help - Solving 2 Trigo Identities

  1. #1
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    Solving 2 Trigo Identities

    hi guys !

    have this 2 trigo identities which i am currently stuck with.

    Appreciate if anyone can enlighten me thanks !
    Attached Thumbnails Attached Thumbnails Solving 2 Trigo Identities-add-maths-question.jpg  
    Last edited by liukawa; November 14th 2010 at 12:27 AM. Reason: attachment cant open
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  2. #2
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    Quote Originally Posted by liukawa View Post
    hi guys !

    have this 2 trigo identities which i am currently stuck with.

    Appreciate if anyone can enlighten me thanks !Attachment 19697
    I'm sorry but I get the message "Invalid Attachment specified." when I click on the attachment.
    Maybe you could try re-uploading it?
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  3. #3
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    hey dude.

    i reposted !

    thanks for informing me about the glitch !
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  4. #4
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    Hello, liukawa!

    \text{Prove that: }\;2\csc^2\!2x - \csc^2\!x \;=\;-2\cot2x\csc2x

    The left side is: . \dfrac{2}{\sin^2\!2x} - \dfrac{1}{\sin^2\!x} \;\;=\;\;\dfrac{2}{(2\sin x\cos x)^2} - \dfrac{1}{\sin^2\!x}


    . . . =\;\;\dfrac{2}{4\sin^2\!x\cos^2\!x} - \dfrac{1}{\sin^2\!x} \;\;=\;\;\dfrac{2}{4\sin^2\!x\cos^2\!x} - \dfrac{4\cos^2\!x}{4\sin^2\!x\cos^2\!x}


    . . . \displaystyle =\;\;\frac{2-4\cos^2\!x}{4\sin^2\!x\cos^2\!x} \;\;=\;\;\frac{2 - 4\left(\frac{1+\cos2x}{2}\right)}{\sin^22x} \;\;=\;\;\frac{2 - 2 - 2\cos2x}{\sin^22x}


    . . . \displaystyle =\; \frac{-2\cos2x}{\sin^22x} \;=\; -2\cdot\frac{\cos2x}{\sin2x}\cdot\frac{1}{\sin2x} \;=\;-2\cot2x\csc2x

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, liukawa!


    The left side is: . \dfrac{2}{\sin^2\!2x} - \dfrac{1}{\sin^2\!x} \;\;=\;\;\dfrac{2}{(2\sin x\cos x)^2} - \dfrac{1}{\sin^2\!x}


    . . . =\;\;\dfrac{2}{4\sin^2\!x\cos^2\!x} - \dfrac{1}{\sin^2\!x} \;\;=\;\;\dfrac{2}{4\sin^2\!x\cos^2\!x} - \dfrac{4\cos^2\!x}{4\sin^2\!x\cos^2\!x}


    . . . \displaystyle =\;\;\frac{2-4\cos^2\!x}{4\sin^2\!x\cos^2\!x} \;\;=\;\;\frac{2 - 4\left(\frac{1+\cos2x}{2}\right)}{\sin^22x} \;\;=\;\;\frac{2 - 2 - 2\cos2x}{\sin^22x}


    . . . \displaystyle =\; \frac{-2\cos2x}{\sin^22x} \;=\; -2\cdot\frac{\cos2x}{\sin2x}\cdot\frac{1}{\sin2x} \;=\;-2\cot2x\csc2x

    Thanks dude !

    You are awesome !

    Any tips on how to solve the 2nd trigo identity?

    Have been cracking my head whole day long and havent got a single clue !

    Appreciate your help greatly thanks !
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