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Math Help - T-Substitution [tan(0.5x)]

  1. #1
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    T-Substitution [tan(0.5x)]

    Can someone show me how using the t=\tan \left(\frac12 x\right) substitution, you can establish:
    \sin x = \frac{2t}{1+t^2}

    \cos x = \frac{1-t^2}{1+t^2}

    It's memorised in my mind (the subsitution) but I don't know the method which bought these substituion in the first place. Thanks in advance.
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  2. #2
    Super Member wingless's Avatar
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    \sin x = \frac{2t}{1+t^2} = \frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}} = \frac{2 \tan \frac{x}{2}}{\sec^2 \frac{x}{2}} = 2 \sin \frac{x}{2} \cos \frac{x}{2} = \sin x

    \cos x = \frac{1-t^2}{1+t^2} = \frac{1- \tan^2 \frac{x}{2}}{1+ \tan^2 \frac{x}{2}} = \frac{\frac{\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2}}}{\sec^2 \frac{x}{2}} = \cos^2 \frac{x}{2}-\sin^2 \frac{x}{2} = \cos x
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  3. #3
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    I know how to go from the substitution to prove it but how would I go about finding the substitution for \sin x and \cos x if I was just given t=\tan \left(\frac12 x\right)

    If I'm just given t=\tan \left(\frac12 x\right) and told to find what \sin x and \cos x would be (in terms of t), how would I find it's substitution?
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  4. #4
    Super Member wingless's Avatar
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    You can see a basic way to find them here.
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  5. #5
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    Refer to the attached image.

    Bobak

    Edit: sorry about my funny looking multiplication sign.
    Attached Thumbnails Attached Thumbnails T-Substitution [tan(0.5x)]-photo.jpg  
    Last edited by bobak; August 2nd 2008 at 05:13 PM.
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  6. #6
    Super Member wingless's Avatar
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    Quote Originally Posted by bobak View Post
    Refer to the attached image.

    Bobak
    You wrote " 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} = 2 \frac{t}{\sqrt{t^2+1}} \color{red}+\color{black}\frac{1}{\sqrt{t^2+1}}". I think that's a typo.
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