# How do you simplify the expression sin (arctan (1/4) + arccos (3/4) )?

$\sin \left(\arctan \left(\frac{1}{4}\right) + \arccos \left(\frac{3}{4}\right)\right) = \frac{3 \sqrt{17} + 4 \sqrt{119}}{68}$

#### Explanation:

Let $A = \arctan \left(\frac{1}{4}\right)$
Let $B = \arccos \left(\frac{3}{4}\right)$

$\sin \left(A + B\right) = \sin A \cos B + \cos A \sin B$

$\sin \left(A + B\right) = \left(\frac{1}{\sqrt{17}}\right) \left(\frac{3}{4}\right) + \left(\frac{4}{\sqrt{17}}\right) \left(\frac{\sqrt{7}}{4}\right)$

$\sin \left(A + B\right) = \frac{3 + 4 \sqrt{7}}{4 \sqrt{17}}$

$\sin \left(A + B\right) = \frac{3 \sqrt{17} + 4 \sqrt{119}}{68}$

$\sin \left(\arctan \left(\frac{1}{4}\right) + \arccos \left(\frac{3}{4}\right)\right) = \frac{3 \sqrt{17} + 4 \sqrt{119}}{68}$

God bless....I hope the explanation is useful.