Math Help - Inverse Trig functions

1. Inverse Trig functions

I have no idea how to simplify these...answers are in the back of my packet but I have no idea how they got them...

cot(arctan(x))

sin(arctan(x))

tan(arccos(x/2))

No calculator, though I don't know why you'd use one ne ways.

Thanks so much!!!

2. Originally Posted by Aurora_Starcrystal
I have no idea how to simplify these...answers are in the back of my packet but I have no idea how they got them...

cot(arctan(x))

sin(arctan(x))

tan(arccos(x/2))

No calculator, though I don't know why you'd use one ne ways.

Thanks so much!!!
I will do the first. the others are similar.

Now we could recall that cot(x) = 1/tan(x)

therefore, cot(arctan(x)) = 1/[tan(arctan(x))] = 1/x ........since the tan undoes the arctan

but here is a general method of how to approach these problems when they don't work out so nicely.

We wish to find $\cot ( \arctan x )$

Let $\arctan x = \theta$

$\Rightarrow x = \tan \theta$

but the tangent ratio is given by the ratio of the side opposite the angle to the side adjacent to the angle, thus we can construct the triangle you see below. We draw the angle theta and fill in the opposite and adjacent side accordingly. We can find the hypotenuse using Pythagoras' theorem, but that was not needed here.

Now, $\cot ( \arctan x ) = \cot \theta = \frac { \mbox { Adjacent }}{ \mbox { Opposite }} = \frac {1}{x}$

Now you try the others

3. Take a gander at the triangles.

the left one is $sin(tan^{-1}(x))$. The right one is

$tan(cos^{-1}(\frac{x}{2}))=\frac{\sqrt{4-x^{2}}}{x}$