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Math Help - identity problems

  1. #1
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    Post identity problems

    Couple of problems from todays homework that i'm struggling with. Thanks for having a look.

    <br />
\frac{sinx+tanx} {1+cos}=\frac{sin2x}{2cos^2x}<br />

    <br />
\frac{sin2x}{2-2cos^2x}=cotx<br />
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  2. #2
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    Quote Originally Posted by nightrider456 View Post
    Couple of problems from todays homework that i'm struggling with. Thanks for having a look.

    <br />
\frac{sinx+tanx} {1+cos}=\frac{sin2x}{2cos^2x}<br />

    <br />
\frac{sin2x}{2-2cos^2x}=cotx<br />
    \frac{\sin{x} + \tan{x}}{1 + \cos{x}} = \frac{\sin{x} + \frac{\sin{x}}{\cos{x}}}{1 + \cos{x}}

     = \frac{\frac{\sin{x}\cos{x} + \sin{x}}{\cos{x}}}{1 + \cos{x}}

     = \frac{\sin{x}\cos{x} + \sin{x}}{\cos{x}(1 + \cos{x})}

     = \frac{\sin{x}(1 + \cos{x})}{\cos{x}(1 + \cos{x})}

     = \frac{\sin{x}}{\cos{x}}

     = \frac{\sin{x}\cos{x}}{\cos^2{x}}

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

     = \frac{\sin{2x}}{2\cos^2{x}}.
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  3. #3
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    Quote Originally Posted by nightrider456 View Post
    Couple of problems from todays homework that i'm struggling with. Thanks for having a look.

    <br />
\frac{sinx+tanx} {1+cos}=\frac{sin2x}{2cos^2x}<br />

    <br />
\frac{sin2x}{2-2cos^2x}=cotx<br />
    \frac{\sin{2x}}{2 - 2\cos^2{x}} = \frac{2\sin{x}\cos{x}}{2 - 2\cos^2{x}}

     = \frac{2\sin{x}\cos{x}}{2(1 - \cos^2{x})}

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

     = \frac{\sin{x}\cos{x}}{\sin^2{x}}

     = \frac{\cos{x}}{\sin{x}}

     = \cot{x}
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