How do you find the exact value of #\cos(tan^{-1}sqrt{3})#?

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Answer 1 · 2014-11-12T15:49:36

Method 1

Let #theta=tan^{-1}sqrt{3}#.

By rewriting in terms of tangent,

#Leftrightarrow tan theta=sqrt{3}={("Opposite")}/{("Adjacent")}#

So, we can let

#("Opposite")=sqrt{3}# and #("Adjacent")=1#.

By Pythagorean Theorem,

#("Hypotenuse")=sqrt{(sqrt{3})^2+1^2}=sqrt{4}=2#.

Hence,

#cos(tan^{-1}sqrt{3})=cos theta={("Adjacent")}/{("Hypotenuse")}=1/2#

Method 2

#cos(tan^{-1}sqrt{3})=cos(pi/3)=1/2#

I hope that this was helpful