# How do you simplify sin(arcsinx + arccosx)?

Sep 3, 2016

$\sin \left(\arcsin x + \arccos x\right) = 1$

#### Explanation:

We can offer two approaches to this problem - using trigonometry of right triangles (applicable for positive $x$) and purely trigonometric (applicable for all $x$, but we will use it for $x \le 0$.

Let's analyze this problem from the position of trigonometry of the right triangle. For this, let's assume that $0 < x \le 1$.

Then $\angle A = \arcsin x$ is an angle from $0$ to $\frac{\pi}{2}$, sine of which is $x$.
Assume, this angle $\angle A$ is an acute angle of a right triangle $\Delta A B C$. So, $\sin \left(\angle A\right)$ is a ratio of an opposite cathetus $a$ to hypotenuse $c$:
$x = \sin \left(\angle A\right) = \frac{a}{c}$

Consider another acute angle of this triangle - $\angle B$.
Expression $\frac{a}{c}$ represents its cosine.
Therefore,
$\angle B = \arccos \left(\frac{a}{c}\right) = \arccos x$

Now we see that $\angle A = \arcsin x$ and $\angle B = \arccos e x$ are two acute angles in a right triangle. Their sum is always $\frac{\pi}{2}$.
Therefore,
$\sin \left(\arcsin x + \arccos x\right) = \sin \left(\frac{\pi}{2}\right) = 1$

Case with non-positive $x$ we will consider differently.
If $x \le 0$, $\arcsin x$ is between $- \frac{\pi}{2}$ and $0$.
If $x \le 0$, $\arccos x$ is between $\frac{\pi}{2}$ and $\pi$.

Using formula for $\sin$ of a sum of two angles,
$\sin \left(\arcsin x + \arccos x\right) =$
$= \sin \left(\arcsin x\right) \cdot \cos \left(\arccos x\right) + \cos \left(\arcsin x\right) \cdot \sin \left(\arccos y\right)$

By definition of inverse trigonometric functions $\arcsin$ and $\arccos$, we write the following:
$\sin \left(\arcsin x\right) = x$
$\cos \left(\arccos x\right) = x$

Using trigonometric identity ${\sin}^{2} \left(\phi\right) + {\cos}^{2} \left(\phi\right) = 1$, we can find the other components:
${\cos}^{2} \left(\arcsin x\right) = 1 - {\sin}^{2} \left(\arcsin x\right) = 1 - {x}^{2}$
${\sin}^{2} \left(\arccos x\right) = 1 - {\cos}^{2} \left(\arccos x\right) = 1 - {x}^{2}$

Since $\arcsin x$ is between $- \frac{\pi}{2}$ and $0$, its cosin is positive:
${\cos}^{2} \left(\arcsin x\right) = 1 - {x}^{2}$
$\implies \cos \left(\arcsin x\right) \sqrt{1 - {x}^{2}}$

Since $\arccos x$ is between $\frac{\pi}{2}$ and $\pi$, its sine is positive:
${\sin}^{2} \left(\arccos x\right) = 1 - {x}^{2}$
$\implies \sin \left(\arccos x\right) \sqrt{1 - {x}^{2}}$

Now we can calculate the value of the original expression:
$\sin \left(\arcsin x + \arccos x\right) =$
$= \sin \left(\arcsin x\right) \cdot \cos \left(\arccos x\right) + \cos \left(\arcsin x\right) \cdot \sin \left(\arccos y\right) =$
$= x \cdot x + \sqrt{1 - {x}^{2}} \cdot \sqrt{1 - {x}^{2}} =$
$= {x}^{2} + 1 - {x}^{2} = 1$
(as in the case we did geometrically).