evaluating inverse trig functions

• October 16th 2012, 12:40 PM
kingsolomonsgrave
evaluating inverse trig functions
evaluate cos(2 tan^-1(1/3))

I figure if I can evaluate arctan of 1/3 I can solve it (multiply by two and take the cosine of the result).

Is it advisable to do this with a calculator or is the value of acrtan(1/3) commonly known value?

thanks!
• October 16th 2012, 01:13 PM
Soroban
Re: evaluating inverse trig functions
Hello, kingsolomonsgrave!

Quote:

$\text{Evaluate: }\:\cos\left[2\tan^{\text{-}1}\!\left(\tfrac{1}{3}\right)\right]$

$\text{Let }\theta \:=\:\tan^{\text{-}1}\!\left(\tfrac{1}{3}\right)$

$\text{Then: }\:\tan\theta \:=\:\frac{1}{3} \:=\:\frac{opp}{adj}$

$\theta\text{ is in a right triangle with: }\,opp = 1,\;adj = 3$
. . $\text{Pythagorus says: }\:hyp = \sqrt{10}$
$\text{Hence: }\:\sin\theta = \tfrac{1}{\sqrt{10}},\;\cos\theta = \tfrac{3}{\sqrt{10}}$

$\text{Therefore: }\:\cos\left[2\tan^{\text{-}1}\!\left(\tfrac{1}{3}\right)\right] \;=\;\cos(2\theta) \;=\;\cos^2\!\theta - \sin^2\!\theta$

. . . . . . . . . . $=\;\left(\frac{3}{\sqrt{10}}\right)^2 - \left(\frac{1}{\sqrt{10}}\right)^2 \;=\;\frac{9}{10} - \frac{1}{10} \;=\;\frac{8}{10} \;=\;\boxed{\frac{4}{5}}$
• October 16th 2012, 01:21 PM
TheEmptySet
Re: evaluating inverse trig functions
Quote:

Originally Posted by kingsolomonsgrave
evaluate cos(2 tan^-1(1/3))

I figure if I can evaluate arctan of 1/3 I can solve it (multiply by two and take the cosine of the result).

Is it advisable to do this with a calculator or is the value of acrtan(1/3) commonly known value?

thanks!

You don't need to know the value. Remember that inverse trig functions are angles!

Let $\alpha = \tan^{-1}\left( \frac{1}{3}\right)$

So by the double angle identity we have

$\cos(2\alpha)=2\cos^2(\alpha)-1$

Now we just need to figure out

$\cos(\tan\left( \frac{1}{3}\right)$

So we can draw a triangle and use the pythagorean theorem

Attachment 25253

This shows that

$\cos(\tan\left( \frac{1}{3}\right)=\frac{3}{\sqrt{10}}$

Now just plug this in to finish..