Answer
$\dfrac{\sqrt 3 -1}{2}$
Work Step by Step
Recall the identity: $(f- g) (x)=f(x) - g(x)$
Let us consider:
$f(x)=\sin x $ and $g(x)=\cos x$
Now, $(f+g) (60^{\circ})=f (60^{\circ})+g(60^{\circ})$
We know from the unit circle that $\sin (60^{\circ}) =\dfrac{\sqrt {3}}{2}$ and $\cos (60^{\circ})=\dfrac{1}{2}$
Thus, we have: $(f -g) (60^{\circ}) =\sin (60^{\circ}) -\cos ((60^{\circ}) \\=\dfrac{\sqrt {3}}{2} - \dfrac{1}{2} \\=\dfrac{\sqrt 3 -1}{2}$
