Answer
$sin(sin^{-1}\frac{1}{2}+tan^{-1}(-3)) = \frac{\sqrt{10}~-~3\sqrt{30}}{20}$
Work Step by Step
Let $A = sin^{-1}\frac{1}{2}$
Then:
$sin~A = \frac{1}{2}$
$cos~A = \frac{\sqrt{2^2-1^2}}{2} = \frac{\sqrt{3}}{2}$
Let $B = tan^{-1}(-3)$
$cos~B = \frac{1}{\sqrt{(-3)^2+1^2}} = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}$
$sin~B = \frac{-3}{\sqrt{(-3)^2+1^2}} = -\frac{3}{\sqrt{10}} = -\frac{3\sqrt{10}}{10}$
We can find $~~sin(sin^{-1}\frac{1}{2}+tan^{-1}(-3))$:
$sin(A+B) = sin~A~cos~B+cos~A~sin~B$
$sin(A+B) = (\frac{1}{2})~(\frac{\sqrt{10}}{10})+(\frac{\sqrt{3}}{2})~(-\frac{3\sqrt{10}}{10})$
$sin(A+B) = \frac{\sqrt{10}}{20}-\frac{3\sqrt{30}}{20}$
$sin(A+B) = \frac{\sqrt{10}~-~3\sqrt{30}}{20}$
Therefore, $~~sin(sin^{-1}\frac{1}{2}+tan^{-1}(-3)) = \frac{\sqrt{10}~-~3\sqrt{30}}{20}$
