y=arctan(x) , show that....

**Twig** · June 11th 2009, 01:47 AM

Hi

I am not sure about my solution, if it´s done "in the right way". Anyway, here it comes:

$\mbox{ Let } y=arctan(x) \, \mbox{ , show that } cos^{2}(y)=\frac{1}{1+x^{2}}$

I did:

$tan(y)=x \Longleftrightarrow \frac{sin(y)}{cos(y)}=x \Rightarrow cos(y)=\frac{sin(y)}{x}$

Now raising both sides to the second power...

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

Thanks!

**yeongil** · June 11th 2009, 01:56 AM

Given: $y = arctan(x)$
Show that: $\cos^{2}(y)=\frac{1}{1+x^{2}}$

Looks right to me. But here's another way, just in case:
Draw a right triangle. Label one of the acute angles y. Since
$y = arctan(x)$ ,
$\tan y = x = \frac{x}{1}$ .
Label the opposite side x and the adjacent side 1. Find the hypotenuse (s) by using the Pythagorean theorem:
$\begin{aligned}<br /> s^2 &= 1^2 + x^2 \\<br /> s &= \sqrt{1 + x^2}<br /> \end{aligned}$
Cosine is adjacent over hypotenuse, so
$\cos y = \frac{1}{\sqrt{1 + x^2}}$ .
Squaring both sides gives you
$\cos^2 y = \frac{1}{1 + x^2}$ .

01

**Soroban** · June 11th 2009, 05:07 AM

Hello, Twig!

Your work looks fine to me . . .

Let: $y\:=\:\arctan x$ , show that: . $\cos^2\!y\:=\:\frac{1}{1+x^2}$

Here's my approach . . .

We have: . $y \:=\:\arctan x\quad\Rightarrow\quad \tan y \:=\:x$

Square both sides: . $\tan^2\!y \:=\:x^2$

$\text{Add 1 to both sides: }\;\underbrace{1 + \tan^2\!y}_{\text{This is }\sec^2\!y} \:=\:1+x^2 \quad\Rightarrow\quad \sec^2\!y \:=\:1+x^2$

Take reciprocals: . $\cos^2\!y \:=\:\frac{1}{1+x^2}$

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