# Thread: y=arctan(x) , show that....

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

Hi

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

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

I did:

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

Now raising both sides to the second power...

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

Thanks!

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

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

01

3. Hello, Twig!

Your work looks fine to me . . .

Let: $\displaystyle y\:=\:\arctan x$, show that: .$\displaystyle \cos^2\!y\:=\:\frac{1}{1+x^2}$
Here's my approach . . .

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

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

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

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