# How do you find the exact value of sin (arcsin(2/3) + arccos(1/3))?

Mar 20, 2018

$\sin \left(\arcsin \left(\frac{2}{3}\right) + \arccos \left(\frac{1}{3}\right)\right) = \frac{2 \left(1 + \sqrt{10}\right)}{9}$

#### Explanation:

Using the formula for the sine of the sum of two angles:

$\sin \left(\arcsin \left(\frac{2}{3}\right) + \arccos \left(\frac{1}{3}\right)\right) = \sin \left(\arcsin \left(\frac{2}{3}\right)\right) \cos \left(\arccos \left(\frac{1}{3}\right)\right) + \cos \left(\arcsin \left(\frac{2}{3}\right)\right) \sin \left(\arccos \left(\frac{1}{3}\right)\right)$

Now, clearly:

$\sin \left(\arcsin \left(\frac{2}{3}\right)\right) = \frac{2}{3}$

$\cos \left(\arccos \left(\frac{1}{3}\right)\right) = \frac{1}{3}$

while:

$\cos \left(\arcsin \left(\frac{2}{3}\right)\right) = \sqrt{1 - {\sin}^{2} \left(\arcsin \left(\frac{2}{3}\right)\right)} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$

$\sin \left(\arccos \left(\frac{1}{3}\right)\right) = \sqrt{1 - {\cos}^{2} \left(\arccos \left(\frac{1}{3}\right)\right)} = \sqrt{1 - \frac{1}{9}} = \frac{2 \sqrt{2}}{3}$

Where we take the positive value because the angles are in the first quadrant.

Then:

$\sin \left(\arcsin \left(\frac{2}{3}\right) + \arccos \left(\frac{1}{3}\right)\right) = \frac{2}{3} \cdot \frac{1}{3} + \frac{\sqrt{5}}{3} \cdot \frac{2 \sqrt{2}}{3} = \frac{2 \left(1 + \sqrt{10}\right)}{9}$