Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've tried following this way, but I haven't succeeded.

Thank you!

share|improve this question
Please add the condition: $x$ is real, since this equation is false in general for complex numbers $x$. –  GEdgar Dec 9 '12 at 14:14
add comment

6 Answers

up vote 2 down vote accepted

Calculate the derivative of both sides:

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


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

Since both derivatives are equal the functions are the same up to the sum of a constant:

$$\arctan x=\arcsin\frac{x}{\sqrt{1-x^2}}+C\,\,\,,\,\,C=\,\text{a constant}$$

Finnaly, to find what $\,C\,$ is you can, for example, input $\,x=0\,$ in the above...

share|improve this answer
This is fine, but I think drawing the picture that gives you the answer without calculus is something that should be known by anyone who asks a question like this. See my answer below. –  Michael Hardy Dec 9 '12 at 17:42
I know, @MichaelHardy. In fact, your answer is exactly the same as amr's, which I upvoted at once and which I'd have accepted as the best one. Perhaps the OP is now in calculus I precisely seeing derivatives and stuff and my answer appealed to him better... –  DonAntonio Dec 9 '12 at 18:18
add comment

Consider the right angled triangle with sides $1,x,\sqrt{1+x^2}$

Let $\phi$ be the angle opposite to the side of length $x$.

We find that: $$\phi=\arcsin(x/\sqrt{1+x^2})$$ $$\phi=\arctan(x/1)$$

Thus: $$\arcsin(x/\sqrt{1+x^2})=\arctan(x)$$

share|improve this answer
+1 Simple, short, elegant...I liked it. –  DonAntonio Dec 9 '12 at 15:18
add comment

put $$x=\tan(\theta)$$ Now rewrite the formula in $\theta$ instead of $x$. All you need, really, are these: $$\tan(x)=\sin(x)/\cos(x)$$ $$\sin^2(x)+\cos^2(x)=1$$ should I be more explicit?

share|improve this answer
Yes, please.... –  matal Dec 9 '12 at 14:02
nameless just did... –  mousomer Dec 9 '12 at 14:17
add comment

Let $\arctan x=y\Leftrightarrow x=\tan y$. Then, $$\sin^2 y+\cos^2 y=1\Leftrightarrow \tan^2 y+1=\frac{1}{\cos^2 y}\Leftrightarrow \frac{1}{x^2+1}=1-\sin^2 y\Leftrightarrow \sin^2 y=\frac{x^2}{x^2+1}$$ and so $$\sin y= \frac{x}{\sqrt{1+x^2}}\Rightarrow \arctan x=y=\arcsin \frac{x}{\sqrt{1+x^2}}$$

share|improve this answer
add comment

Let $\displaystyle\arctan x= y$

$\implies(i) \tan y =x$

and $(ii)\displaystyle-\frac\pi2\le y\le\frac\pi2$ (using the definition of principal value)

$\implies \cos y\ge0$

We have $$\frac{\sin y}x=\frac {\cos y }1=\pm\sqrt{\frac{\sin^2y+\cos^2y}{x^2+1^2}}=\pm\frac1{\sqrt{x^2+1}}$$

$\displaystyle\implies \cos y=+\frac1{\sqrt{x^2+1}}$ and $\displaystyle\sin y=\frac x{\sqrt{x^2+1}}$

So, $\displaystyle\arctan x= y=\arcsin\frac x{\sqrt{x^2+1}}=\arccos\frac1{\sqrt{x^2+1}}$

share|improve this answer
add comment

As soon as you see $\arctan x$, draw a right triangle in which the "opposite" side has length $x$ and the "adjacent" side has length $1$. Then the angle to which those are "opposite" and "adjacent" is $\arctan x$.

The Pythagorean theorem then tells you the length of the hypotenuse.

That gives you the sine of the angle, since $\sin=\dfrac{\mathrm{opp}}{\mathrm{hyp}}$.

That tells you what the angle in question is the arcsine of.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.