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    If  \cos \theta = \frac{1-t^2}{1+t^2} does  \cos \frac{1}{2} \theta = \frac{4-t^2}{4+t^2} i.e. can i just half the t if i half the  \theta ? Because I think it is probably wrong.
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    Think about it:

     1 + t^2 = \sec^2\frac{\theta}{2}

    Thus,  \frac{1}{1+t^2} = \cos^2\frac{\theta}{2}

     \cos\frac{\theta}{2} = \sqrt\frac{1}{1+t^2}

    If you want to see a lucid example of the error remember that  \tan\theta = \frac{2t}{1-t^2} . According to your example above, by halving  \theta :  \tan\frac{\theta}{2} = \frac{4t}{4-t^2} But, the defining property of the t-substitution is  \tan\frac{\theta}{2} = t .

    Hope that makes sense, all the best.
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    (Original post by ziu)
    Think about it:

     1 + t^2 = \sec^2\frac{\theta}{2}

    Thus,  \frac{1}{1+t^2} = \cos^2\frac{\theta}{2}

     \cos\frac{\theta}{2} = \sqrt\frac{1}{1+t^2}

    If you want to see a lucid example of the error remember that  \tan\theta = \frac{2t}{1-t^2} . According to your example above, by halving  \theta :  \tan\frac{\theta}{2} = \frac{4t}{4-t^2} But, the defining property of the t-substitution is  \tan\frac{\theta}{2} = t .

    Hope that makes sense, all the best.
    Thanks, all I had to do was use the equation  \cos 2x = 2 \cos^2 x - 1 and sub  \theta = 2x for it to work!
 
 
 
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