# How do you simplify the expression Sin(arctan(x)+arccos(x))?

Jul 26, 2016

sin 2x

#### Explanation:

arctan x --> x
arccos x --> x
sin (arctan x + arccos x) = sin (x + x) = sin 2x

Jul 26, 2016

.$= \left(\frac{1}{\sqrt{1 + {x}^{2}}}\right) \left(\sqrt{1 - {x}^{2}} \pm {x}^{2}\right)$, x in [-1, 1].

The negative sign is used, when $x \in \left[- 1 , 0\right]$.

#### Explanation:

Let a = arc tan (x). The principal value of $a \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

Then x = tan a. sin a = $\pm \frac{x}{\sqrt{1 + {x}^{2}}}$ and cos a = $\frac{1}{\sqrt{1 + {x}^{2}}}$.

Let b = arc cos x. The principal value of $b \in \left[0 , \pi\right]$

Then, x = cos b and sin b = $\sqrt{1 - {x}^{2}}$. Also, $x \in \left[- 1 , 1\right]$.

The given expression =

$\sin \left(a + b\right)$

$= \sin a \cos b + \cos a \sin b$

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

$= \left(\frac{1}{\sqrt{1 + {x}^{2}}}\right) \left(\sqrt{1 - {x}^{2}} \pm {x}^{2}\right)$, x in [-1, 1].

The negative sign is used when $x \in \left[- 1 , 0\right]$.