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Math Help - more trig questions

  1. #1
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    more trig questions

    Hi
    I am having trouble solving the following:

    1)show that \frac{1+sin{2\theta}+cos{2\theta}}{1+sin{2\theta}-cos{2\theta}}=cot{\theta}

    2)Given t=tan{\frac{x}{2}}, prove that sin(x)=\frac{2t}{1+t^2}
    Hence solve the equation 2sin(x)-cos(x)=1

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I am having trouble solving the following:

    1)show that \frac{1+sin{2\theta}+cos{2\theta}}{1+sin{2\theta}-cos{2\theta}}=cot{\theta}

    2)Given t=tan{\frac{x}{2}}, prove that sin(x)=\frac{2t}{1+t^2}
    Hence solve the equation 2sin(x)-cos(x)=1

    P.S
    1) \frac{1 + \sin{2\theta} + \cos{2\theta}}{1 + \sin{2\theta} - \cos{2\theta}} = \frac{1 + 2\sin{\theta}\cos{\theta} + \cos^2{\theta} - \sin^2{\theta}}{1 + 2\sin{\theta}\cos{\theta} - (\cos^2{\theta} - \sin^2{\theta})}

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

     = \frac{2\cos{\theta}(\sin{\theta} + \cos{\theta})}{2\sin{\theta}(\sin{\theta} + \cos{\theta})}

     = \frac{\cos{\theta}}{\sin{\theta}}

     = \cot{\theta}.
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  3. #3
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I am having trouble solving the following:

    1)show that \frac{1+sin{2\theta}+cos{2\theta}}{1+sin{2\theta}-cos{2\theta}}=cot{\theta}

    2)Given t=tan{\frac{x}{2}}, prove that sin(x)=\frac{2t}{1+t^2}
    Hence solve the equation 2sin(x)-cos(x)=1

    P.S
    If \sin{2\theta} = 2\sin{\theta}\cos{\theta}

     = 2\left(\frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}}\right)\left(\frac{1}{\sqrt{1 + \tan^2{\theta}}}\right)

     = \frac{2\tan{\theta}}{1 + \tan^2{\theta}}.


    Now let \theta = \frac{x}{2}.


    Now to solve 2\sin{x} - \cos{x} = 1

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

    \frac{4t}{1 + t^2} - \sqrt{1 - \left(\frac{2t}{1 + t^2}\right)^2} = 1

    \frac{4t}{1 + t^2} - \sqrt{\frac{(1 + t^2)^2 - 4t^2}{(1 + t^2)^2}} = 1

    \frac{4t}{1 + t^2} - \sqrt{\frac{1 + 2t^2 + t^4 - 4t^2}{(1 + t^2)^2}} = 1

    \frac{4t}{1 + t^2} - \sqrt{\frac{1 - 2t^2 + t^4}{(1 + t^2)^2}} = 1

    \frac{4t}{1 + t^2} - \sqrt{\frac{(1 - t^2)^2}{(1 + t^2)^2}} = 1

    \frac{4t}{1 + t^2} - \frac{1 - t^2}{1 + t^2} = 1

    \frac{t^2 + 4t - 1}{1 + t^2} = 1

    t^2 + 4t - 1 = 1 + t^2

    4t = 2

    t = \frac{1}{2}.


    Now let t = \tan{\frac{x}{2}} and solve for x.
    Last edited by Prove It; January 7th 2010 at 04:25 PM.
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  4. #4
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    can you explain to me how you got this <br />
2\cos^2{\theta} - 1 and 1 - 2\sin^2{\theta}
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  5. #5
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    The Pythagorean Identity.

    \sin^2{\theta} + \cos^2{\theta} = 1

    So \sin^2{\theta} = 1 - \cos^2{\theta} and \cos^2{\theta} = 1 - \sin^2{\theta}.
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  6. #6
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    ok thanks i understand now.
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  7. #7
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    can you explain to me how you got this? <br />
= 2\left(\frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}}\right)\left(\frac{1}{\sqrt{1 + \tan^2{\theta}}}\right)<br />
    Last edited by Paymemoney; January 7th 2010 at 05:07 PM.
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  8. #8
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    \sin{\theta} = \frac{\tan{\theta}}{\sqrt{1 + \tan^2{\theta}}} and \cos{\theta} = \frac{1}{\sqrt{1 + \tan^2{\theta}}}.

    These are also easily proven using the Pythagorean Identity.
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