Sine of Arcsine Arctangent Sum

Let $\alpha = \sin^{-1} \left(\frac{3}{5}\right)$ and $\beta = \tan^{-1} (2)$ . Since $\sin \alpha = \sin\left(\sin^{-1} \left(\frac{3}{5}\right)\right) = \frac{3}{5}$ , thus we can form a right triangle with $5$ as the hypotenuse and $3$ as a side length opposite to angle $\alpha$ . By the Pythagorean theorem, the side length adjacent to angle $\alpha$ is $\sqrt{5^2-3^2} = 4$ . Therefore $\cos\alpha = \frac{4}{5}$ .

Similarily for $\beta$ , we can form a right triangle with $2$ as the side length opposite to angle $\beta$ and $1$ as a side length adjacent to angle $\beta$ . By the Pythagorean theorem, the hypotenuse is $\sqrt{1^2+2^2} = \sqrt{5}$ . Therefore $\sin \beta = \frac{2}{\sqrt{5}}$ and $\cos \beta = \frac{1}{\sqrt{5}}$ .

Substituting in the above, we have $\begin{aligned} \sin\left(\sin^{-1} \left(\frac{3}{5}\right) + \tan^{-1} (2)\right) &= \sin(\alpha + \beta) \\ &= \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ &= \frac{3}{5}\cdot \frac{1}{\sqrt{5}} + \frac{4}{5} \cdot \frac{2}{\sqrt{5}} \\ &= \frac{11}{5\sqrt{5}} \\ &= \frac{11\sqrt{5}}{25} \\ \end{aligned}$

Hence $a+b+c = 11 + 5 + 25 = 41$ .