# How do you evaluate sin (arctan (1/4) + arccos (3/4))?

Jul 3, 2016

Multiple values possible but assuming angles $A$ and $B$ in first quadrant, $\sin \left(\arctan \left(\frac{1}{4}\right) + \arccos \left(\frac{3}{4}\right)\right) = \frac{3 + 4 \sqrt{7}}{4 \sqrt{17}}$

#### Explanation:

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

$\tan A = \frac{1}{4}$ and $\cos B = \frac{3}{4}$ and $\sin B = \sqrt{1 - {\left(\frac{3}{4}\right)}^{2}} = \pm \frac{\sqrt{7}}{4}$

$\tan A = \frac{1}{4}$ leads to $\cot A = 4$ and

$\csc A = \pm \sqrt{1 + {\cot}^{2} A} = \pm \sqrt{1 + {4}^{2}} = \pm \sqrt{17}$ and $\sin A = \pm \frac{1}{\sqrt{17}}$ and $\cos A = \pm \frac{4}{\sqrt{17}}$ - note that they will have same sign.

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

= $\sin A \cos B + \cos A \sin B$

Hence assuming all angles in first quadrant,

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