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!
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!
Hello, kingsolomonsgrave!
$\displaystyle \text{Evaluate: }\:\cos\left[2\tan^{\text{-}1}\!\left(\tfrac{1}{3}\right)\right]$
$\displaystyle \text{Let }\theta \:=\:\tan^{\text{-}1}\!\left(\tfrac{1}{3}\right)$
$\displaystyle \text{Then: }\:\tan\theta \:=\:\frac{1}{3} \:=\:\frac{opp}{adj}$
$\displaystyle \theta\text{ is in a right triangle with: }\,opp = 1,\;adj = 3$
. . $\displaystyle \text{Pythagorus says: }\:hyp = \sqrt{10}$
$\displaystyle \text{Hence: }\:\sin\theta = \tfrac{1}{\sqrt{10}},\;\cos\theta = \tfrac{3}{\sqrt{10}}$
$\displaystyle \text{Therefore: }\:\cos\left[2\tan^{\text{-}1}\!\left(\tfrac{1}{3}\right)\right] \;=\;\cos(2\theta) \;=\;\cos^2\!\theta - \sin^2\!\theta$
. . . . . . . . . . $\displaystyle =\;\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}} $
You don't need to know the value. Remember that inverse trig functions are angles!
Let $\displaystyle \alpha = \tan^{-1}\left( \frac{1}{3}\right)$
So by the double angle identity we have
$\displaystyle \cos(2\alpha)=2\cos^2(\alpha)-1$
Now we just need to figure out
$\displaystyle \cos(\tan\left( \frac{1}{3}\right)$
So we can draw a triangle and use the pythagorean theorem
This shows that
$\displaystyle \cos(\tan\left( \frac{1}{3}\right)=\frac{3}{\sqrt{10}}$
Now just plug this in to finish..