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Math Help - Trig Identities

  1. #1
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    Trig Identities

    Hi, I'm having no luck solving a couple of identities, could someone be of assistance?

    1) (1-sin^2x)/(cosx) = sin2x/2sinx
    2) cscx/cosx = tanx + cosx
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  2. #2
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    Quote Originally Posted by millerst View Post
    Hi, I'm having no luck solving a couple of identities, could someone be of assistance?

    1) (1-sin^2x)/(cosx) = sin2x/2sinx
    2) cscx/cosx = tanx + cosx
    hints for #1 ...

    1-\sin^2{x} = \cos^2{x}

    \sin(2x) = 2\sin{x}\cos{x}

    #2 is not an identity
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  3. #3
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    Hello, millerst!

    The second one isn't true . . .


    1)\;\;\frac{1-\sin^2\!x}{\cos x} \:=\:\frac{\sin2x}{2\sin x}

    The left side is: . \frac{\overbrace{1-\sin^2\!x}^{\text{This is }\cos^2\!x}}{\cos x} \;=\;\frac{\cos^2\!x}{\cos x} \;=\;\cos x

    \text{Multiply by }\frac{2\sin x}{2\sin x}\!:\;\;\cos x\cdot\frac{2\sin x}{2\sin x} \;=\;\frac{\overbrace{2\sin x\cos x}^{\text{This is }\sin2x}}{2\sin x} \;=\;\frac{\sin2x}{2\sin x}

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  4. #4
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    Just wondering, I find these things rather hard, and normally I pick of concepts quickly. Is there anything I can do to make these things easier?
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