# simplifying inverse and normal trig functions

• Aug 29th 2007, 09:20 PM
mer1988
simplifying inverse and normal trig functions
i really need to know how to simplify this equation for -infinity<t<infinity,
cos(arctan(t))...

thanks
• Aug 29th 2007, 10:43 PM
red_dog
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, $\cos(\arctan t)=\frac{1}{\sqrt{1+t^2}}$