# Arc- trigonometric functions

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.

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.

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

• Thanks! I just realized how stupid my question looks (being 2nd year Computer Science, lol) Mar 17, 2015 at 14:50
• Can you give me a tip for the 3rd? I'm stuck at transforming it to cosine. Mar 17, 2015 at 15:02
• Easier, $$\sin^2(\arctan x)=1-\cos^2(\arctan x)=1-\frac{1}{1+x^2}=\frac{x^2}{1+x^2}$$ and thus $$|\sin(\arctan x)|=\frac{|x|}{\sqrt{1+x^2}}\implies \sin(\arctan x)=\frac{x}{\sqrt{1+x^2}}$$
– Surb
Mar 17, 2015 at 15:06

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