# Inverse Trig Ratios

• Oct 16th 2011, 04:43 PM
Manni
Inverse Trig Ratios
How would I evaluate cos(2 arctan(2))?
• Oct 16th 2011, 05:40 PM
SammyS
Re: Inverse Trig Ratios
Use $\displaystyle \cos(2\theta)=1-2\sin^2(\theta)\,.$

Then can you find $\displaystyle \sin(\arctan(2))\,?$
• Oct 16th 2011, 05:50 PM
Manni
Re: Inverse Trig Ratios
Do you mind working it out for me?
• Oct 16th 2011, 07:08 PM
Soroban
Re: Inverse Trig Ratios
Hello, Manni!

Quote:

$\displaystyle \text{Evaluate: }\,\cos(2 \arctan 2 )$

Let $\displaystyle \theta \,=\, \arctan 2 \quad\Rightarrow\quad \tan\theta \,=\,2$

Then: .$\displaystyle \tan\theta \:=\:\tfrac{2}{1} \:=\:\tfrac{opp}{adj}$

That is, $\displaystyle \theta$ is in a right triangle with $\displaystyle opp = 2,\;adj = 1.$
. . Hence: .$\displaystyle hyp \,=\,\sqrt{5} \quad\Rightarrow\quad \cos\theta \,=\,\tfrac{adj}{hyp} \,=\,\tfrac{1}{\sqrt{5}}$

We have: .$\displaystyle \cos 2\theta \;=\;2\cos^2\!\theta - 1 \;=\;2\!\left(\tfrac{1}{\sqrt{5}}\right)^2 - 1 \;=\;\tfrac{2}{5} - 1 \;=\;\text{-}\tfrac{3}{5}$

• Oct 16th 2011, 08:13 PM
Manni
Re: Inverse Trig Ratios
Thank you! This clarified everything!