1. ## Equation

Hi people,

I want to show that $\displaystyle sin(Arctanx)=\frac{x}{\sqrt{1+x^2}}$, can you help me please???

2. let $\displaystyle tan^{-1}x=y$

$\displaystyle x=tan y$

$\displaystyle sec^2 y=1+tan^2 y$

$\displaystyle sec^2 y=1+x^2$

$\displaystyle sec y=\sqrt{1+x^2}$

$\displaystyle sin y=tan y cos y$

$\displaystyle sin y=\frac{tan y}{sec y}$

$\displaystyle sin y=\frac{x}{\sqrt{1+x^2}}$

$\displaystyle sin(tan^{-1}x)=\frac{x}{\sqrt{1+x^2}}$

3. Originally Posted by lehder
Hi people,

I want to show that $\displaystyle sin(Arctanx)=\frac{x}{\sqrt{1+x^2}}$, can you help me please???
For a (imho) somewhat more intuitive solution consider a right triangle with acute angle $\displaystyle \alpha$, opposite leg $\displaystyle x$ and adjacent leg 1: By construction we have $\displaystyle \arctan(x)=\alpha$. But by Pythagoras's theorem we also have that the length of the hypotenuse must be $\displaystyle \sqrt{1+x^2}$. By definition of $\displaystyle \sin$ in a right triangle we find that $\displaystyle \sin(\alpha)=\frac{x}{\sqrt{1+x^2}}$, i.e. the ratio of opposite leg to hypotenuse in that right triangle.