Answer
$y_1(\frac{\pi}{6}) = \frac{1}{2}$
$y_2(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$
$(y_1+y_2)(\frac{\pi}{6}) = \frac{1}{2}+\frac{\sqrt{3}}{2}$
We can see that: $~~y_1(\frac{\pi}{6})+y_2(\frac{\pi}{6}) = (y_1+y_2)(\frac{\pi}{6})$
Work Step by Step
$y_1 = sin~x$
$y_2 = sin~2x$
$y_1+y_2 = sin~x+sin~2x$
We can evaluate each function at $\frac{\pi}{6}$:
$y_1(\frac{\pi}{6}) = sin~\frac{\pi}{6} = \frac{1}{2}$
$y_2(\frac{\pi}{6}) = sin~\frac{2\pi}{6} = sin~\frac{\pi}{3} = \frac{\sqrt{3}}{2}$
$(y_1+y_2)(\frac{\pi}{6}) = sin~\frac{\pi}{6}+sin~\frac{2\pi}{6} = \frac{1}{2}+\frac{\sqrt{3}}{2}$
We can see that: $~~y_1(\frac{\pi}{6})+y_2(\frac{\pi}{6}) = (y_1+y_2)(\frac{\pi}{6})$
