Question

16. (Sin3a-sina*cos2x)/(sin3a+sina)

Answer 1

$\frac{sin3a-sina\cdot cos2a}{sin3a+sina}=\frac{sin3a-\frac{1}{2}(sin3a+sin(-a))}{sin3a+sina}=\frac{\frac{1}{2}sin3a+\frac{1}{2}sina}{sin3a+sina} = \\\\= \frac{\frac{1}{2}(sin3a+sina)}{sin3a+sina} =\frac{1}{2}$