Answer
$cos(sin^{-1}\frac{3}{5}+cos^{-1}\frac{5}{13}) = -\frac{16}{65}$
Work Step by Step
Let $A = sin^{-1}\frac{3}{5}$
Then:
$sin~A = \frac{3}{5}$
$cos~A = \frac{\sqrt{5^2-3^2}}{5} = \frac{4}{5}$
Let $B = cos^{-1}\frac{5}{13}$
$cos~B = \frac{5}{13}$
$sin~B = \frac{\sqrt{13^2-5^2}}{13} = \frac{12}{13}$
We can find $~~cos(sin^{-1}\frac{3}{5}+cos^{-1}\frac{5}{13})$:
$cos(A+B) = cos~A~cos~B-sin~A~sin~B$
$cos(A+B) = (\frac{4}{5})~(\frac{5}{13})-(\frac{3}{5})~(\frac{12}{13})$
$cos(A+B) = \frac{20}{65}-\frac{36}{65}$
$cos(A+B) = -\frac{16}{65}$
Therefore, $~~cos(sin^{-1}\frac{3}{5}+cos^{-1}\frac{5}{13}) = -\frac{16}{65}$
