9
$\begingroup$

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

Thank you!

$\endgroup$
1
  • 2
    $\begingroup$ Please add the condition: $x$ is real, since this equation is false in general for complex numbers $x$. $\endgroup$
    – GEdgar
    Dec 9, 2012 at 14:14

7 Answers 7

19
$\begingroup$

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

$\endgroup$
0
8
$\begingroup$

Calculate the derivative of both sides:

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

$$\left(\arcsin\frac{x}{\sqrt{1+x^2}}\right)'=\frac{\sqrt{1+x^2}-\frac{x^2}{\sqrt{1+x^2}}}{1+x^2}\cdot\frac{1}{\sqrt{1-\frac{x^2}{1+x^2}}}=$$

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

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

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

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

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$ Dec 9, 2012 at 17:42
  • $\begingroup$ 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... $\endgroup$
    – DonAntonio
    Dec 9, 2012 at 18:18
5
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $$\sin y= \frac{x}{\sqrt{1+x^2}}\Rightarrow \arctan x=y=\arcsin \frac{x}{\sqrt{1+x^2}}$$

$\endgroup$
0
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Yes, please.... $\endgroup$
    – Alex
    Dec 9, 2012 at 14:02
  • $\begingroup$ nameless just did... $\endgroup$
    – mousomer
    Dec 9, 2012 at 14:17
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

To prove that $$\sin(\arctan x)=\frac{x}{(1+x^2)^{\frac12}}$$ let $a=\arctan x$. Then $$s=\sin a=\tan a\cdot\cos a=\tan a(1-(\sin a)^2)^{\frac12}$$

$$s^2=(x^2)\cdot(1-s^2)$$

$$s^2=\frac{x^2}{1+x^2}$$

$$s=\sin a=\sin a(\arctan x)=\frac{x}{(1+x^2)^{\frac12}}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .