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Math Help - Trig identity

  1. #1
    Senior Member Stroodle's Avatar
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    Trig identity

    Hey there,

    I have this problem that I've been able to solve, but only after what seems like an excessive amount of working. Was wondering if anyone knows a quicker way to go about it?

    Given that t=tan\left (\frac{x}{2}\right ) show that sin(x)=\frac{2t}{1+t^2}


    I've done this by first showing that cos(x)=\frac{1-t^2}{1+t^2}, but I'm sure there must be a shorter way, or something simple that I'm missing?

    THanks
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  2. #2
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    Quote Originally Posted by Stroodle View Post
    Hey there,

    I have this problem that I've been able to solve, but only after what seems like an excessive amount of working. Was wondering if anyone knows a quicker way to go about it?

    Given that t=tan\left (\frac{x}{2}\right ) show that sin(x)=\frac{2t}{1+t^2}


    I've done this by first showing that cos(x)=\frac{1-t^2}{1+t^2}, but I'm sure there must be a shorter way, or something simple that I'm missing?

    THanks
    \frac{2t}{1 + t^2} = \frac{2\tan{\frac{x}{2}}}{1 + \tan^2{\frac{x}{2}}}

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

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

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

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

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


    Now let \theta = \frac{x}{2}, so that this becomes

    2\sin{\theta}\cos{\theta}

     = \sin{2\theta}

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

     = \sin{x}.
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  3. #3
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    Hello, Stroodle!

    Another approach . . .


    \text{Given that: }\,t\:=\:\tan\tfrac{x}{2},\, \text{ show that: }\:\sin x \:=\:\frac{2t}{1+t^2}

    We have: . \tan\tfrac{x}{2} \:=\:\frac{t}{1} \:=\:\frac{opp}{adj}

    Angle \tfrac{x}{2} is in a right triangle with: . opp = t,\;adj = 1

    Pythagorus says: . hyp \,=\,\sqrt{1+t^2}

    Hence: . \sin\tfrac{x}{2} \:=\:\frac{t}{\sqrt{1+t^2}},\;\;\cos\tfrac{x}{2} \:=\:\frac{1}{\sqrt{1+t^2}} .[1]



    \text{We know that: }\;\sin x \;=\;2\sin\tfrac{x}{2}\cos\tfrac{x}{2} .[2]


    Substitute [1] into [2]: . \sin x \;=\;2\left(\frac{t}{\sqrt{1+t^2}}\right)\left(\fr  ac{1}{\sqrt{1+t^2}}\right) \;=\;\frac{2t}{1+t^2}

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