Inverse Trig Ratios

**Manni** · October 16th 2011, 04:43 PM

How would I evaluate cos(2 arctan(2))?

**SammyS** · October 16th 2011, 05:40 PM

Use $\cos(2\theta)=1-2\sin^2(\theta)\,.$

Then can you find $\sin(\arctan(2))\,?$

**Manni** · October 16th 2011, 05:50 PM

Do you mind working it out for me?

**Soroban** · October 16th 2011, 07:08 PM

Hello, Manni!

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

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

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

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

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

**Manni** · October 16th 2011, 08:13 PM

Thank you! This clarified everything!

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