# How do you evaluate Sin(arcsin(3/5)+(arctan2))?

Aug 11, 2016

The reqd. value $= \frac{11 \sqrt{5}}{25}$.

#### Explanation:

Let arcsin(3/5)=alpha, &, arctan2=beta; alpha, beta in (-pi/2,pi/2).

Hence, by defn., $\sin \alpha = \frac{3}{5} > 0 s o , \alpha \in \left(0 , \frac{\pi}{2}\right)$, and,

$\tan \beta = 2 > 0 , s o , \beta \in \left(0 , \frac{\pi}{2}\right)$

Now, reqd. value

$= \sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \ldots \ldots . \left(1\right)$

Therefore, to find this, we will need $\cos \alpha , \sin \beta , \cos \beta$

$\sin \alpha = \frac{3}{5} \Rightarrow \cos \alpha = + \sqrt{1 - {\sin}^{2} \alpha} = + \frac{4}{5} , a s , \alpha \in \left(0 , \frac{\pi}{2}\right)$

$\cos \beta = \frac{1}{\sec} \beta = \frac{1}{+ \sqrt{1 + {\tan}^{2} \beta}} = \frac{1}{+ \sqrt{5}}$, and,

$\sin \beta = \tan \beta \cos \beta = 2 \left(\frac{1}{+ \sqrt{5}}\right) = + \frac{2}{\sqrt{5}} , a s \beta \in \left(0 , \frac{\pi}{2}\right)$

Using all these in $\left(1\right)$, we get,

the reqd. value $= \frac{3}{5} \cdot \frac{1}{\sqrt{5}} + \frac{4}{5} \cdot \frac{2}{\sqrt{5}} = \frac{11}{5 \sqrt{5}} = \frac{11 \sqrt{5}}{25}$.