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Math Help - simplifying inverse and normal trig functions

  1. #1
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    simplifying inverse and normal trig functions

    i really need to know how to simplify this equation for -infinity<t<infinity,
    cos(arctan(t))...

    thanks
    Last edited by mer1988; August 29th 2007 at 10:34 PM.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let \arctan t=\theta\Rightarrow \tan\theta =t,\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)
    So we have to find \cos\theta, knowing \tan\theta.
    We have \displaystyle\cos\theta=\pm\frac{1}{\sqrt{1+\tan^2  \theta}}
    On the interval \left(-\frac{\pi}{2},\frac{\pi}{2}\right), \cos\theta\geq 0, so we choose the sign +
    Thus, \cos(\arctan t)=\frac{1}{\sqrt{1+t^2}}
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