Arc- trigonometric functions

Question

This is, perhaps, too simple of a question for here, but I'd love it if someone helped me out.

I'm just learning about arc- trigonometric functions (because I failed both calculus exams) and my textbook says that it's pretty obvious how to prove the following equations. However, I've been trying for some time now and I can't seem to get the answers right. (Also, couldn't find the same problem anywhere else)

$$\begin{align*} \cos^2(\operatorname{arctg}x) &= \frac{1}{1+x^2}\\\\ \operatorname{tg}(\arcsin x) &= \frac{x}{\sqrt{1-x^2}}\\\\ \sin(\operatorname{arctg} x) &= \frac{x}{\sqrt{1+x^2}}\\\\ \arcsin x &=\pi/2 - \arccos x \end{align*}$$

I don't expect an answer to all of those. I suspect that they are very close to one another. If you would just help me out with the ideas, that would be great as well.

Answer 1

For the first demonstration, use the following variable change $y = \arctan x$, so that $x = \tan y$ and $$ \cos^2 y = \frac{1}{1 + \tan^2 y} $$ Next, by direct substitution of $\tan y = \sin y / \cos y$ and $\sin^2 y + \cos^2 y = 1$ you have the proof. I think that using this variable change the others proof can easily be done.

Answer 2

Hint

Use that

$$\cos^2(x)=\frac{1+\cos(2x)}{2}$$

$$\sin^2(x)=1-\cos^2(x)$$

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

and $$\tan(x)=\frac{\sin x}{\cos x}.$$

Answer 3

Hint: by definition you have: $$ \arctan x=y \iff x=\tan y $$ $$ \arcsin x=y \iff x=\sin y $$ $$ \arccos x=y \iff x=\cos y $$

Using this definitions you can solve your equations. E.g. the first one become:

$$ \cos^2y=\dfrac{1}{1+x^2} $$ and, since $x=\tan y$

$$ \cos^2y=\dfrac{1}{1+\tan^2 y} $$ that you can easely verify.

This way is useful for eqautions with inverse trigonometric functions