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.


3 Answers 3


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.



Use that




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

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


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