How do you find the exact value of \cos(tan^{-1}sqrt{3})cos(tan13)?

1 Answer
Nov 12, 2014

Method 1

Let theta=tan^{-1}sqrt{3}θ=tan13.

By rewriting in terms of tangent,

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

So, we can let

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

By Pythagorean Theorem,

("Hypotenuse")=sqrt{(sqrt{3})^2+1^2}=sqrt{4}=2(Hypotenuse)=(3)2+12=4=2.

Hence,

cos(tan^{-1}sqrt{3})=cos theta={("Adjacent")}/{("Hypotenuse")}=1/2cos(tan13)=cosθ=(Adjacent)(Hypotenuse)=12


Method 2

cos(tan^{-1}sqrt{3})=cos(pi/3)=1/2cos(tan13)=cos(π3)=12


I hope that this was helpful