# how does one evaluate cos (2*arctan of (1/3))?

• Oct 23rd 2012, 09:37 PM
kingsolomonsgrave
how does one evaluate cos (2*arctan of (1/3))?
Attachment 25381

I can do this on my calculator quite easily but I'm not sure how to do it algebratically
• Oct 23rd 2012, 09:55 PM
Soroban
Re: how does one evaluate cos (2*arctan of (1/3))?
Hello, kingsolomonsgrave!

Quote:

$\text{Evaluate: }\:\cos\left(\arctan\tfrac{1}{3}\right)$

$\text{Let }\,\theta \,=\,\arctan\tfrac{1}{3}$

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

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

$\text{Therefore: }\:\cos\left(\arctan\tfrac{1}{3}\right) \;=\;\cos\theta \;=\;\frac{3}{\sqrt{10}}$
• Oct 23rd 2012, 10:00 PM
kingsolomonsgrave
Re: how does one evaluate cos (2*arctan of (1/3))?
thanks!

So in order to work with 2*acrtan, at which stage would I multiply by 2?

would I multiply tan theta by 2 and thus take the cosine of 2/3?

I may need sleep soon, I think my brain is shutting down, this seems like it should be obvious. Sorry if im being dense (Worried)
• Oct 23rd 2012, 10:05 PM
MarkFL
Re: how does one evaluate cos (2*arctan of (1/3))?
Use a double-angle identity for cosine.