
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.

Hi CFem
Question 1
let :
$\displaystyle arc \,\tan \frac{1}{\sqrt{3}} = x$
$\displaystyle \tan x = \frac{1}{\sqrt{3}}$
Find x
Question 2
let :
$\displaystyle arc\, \sec 2 = y$
$\displaystyle \sec y = 2$
$\displaystyle \frac{1}{\cos y}=2$
$\displaystyle \cos y = \frac{1}{2}$
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)
:)