# Inverse Trig functions

• Sep 15th 2009, 04:25 PM
CFem
Inverse Trig functions
I'll just throw the three questions out first:

tan^-1(1/sqrt(3))
sec^-1(2)

Both of those I need the exact answer

cos(2tan^-1x)

That I need simplified.

(this is all supposed to be revision, by the way)

For the first two we were told to draw a triangle, and put the values there to make solving the problem geometrically simple, but I was never really taught the difference between cosine and arcsine (or any of the other ones, for that matter), so I'm not sure how to proceed

The last one I just don't know how to do.
• Sep 15th 2009, 08:23 PM
songoku
Hi CFem

Question 1
let :
$arc \,\tan \frac{1}{\sqrt{3}} = x$

$\tan x = \frac{1}{\sqrt{3}}$

Find x

Question 2
let :
$arc\, \sec 2 = y$

$\sec y = 2$

$\frac{1}{\cos y}=2$

$\cos y = \frac{1}{2}$

Now can you draw a right angle triangle and set the value of adjacent side and hypotenuse, then find y ?

Question 3
let : arc tan (x) = p , so cos(2 [arc tan x] ) = cos (2p)

arc tan (x) = p
tan (p) = x

Now draw a right angle triangle then set the value of opposite and adjacent side, then use double angle formula to simplify cos(2p)

:)