Chapter 5 - Trigonometric Functions - Section 5.2 Trigonometric Functions: Unit Circle Approach - 5.2 Assess Your Understanding - Page 403: 114

$1$

Recall the rule: $(h \circ \ f ) (x) =h[f(x)]$ Let us consider that $f(x)=\sin x $ and $h(x)= 2 x$ We know from the unit circle that $\sin (\dfrac{5 \pi}{6}) =\dfrac{1}{2}$ Thus, we have: $( h \ \circ \ h ) (\dfrac{5 \pi}{6}) =h[f(\dfrac{5 \pi}{6})] \\= h [\sin (\dfrac{5 \pi}{6})] \\=h (\dfrac{1}{2}) \\=(2) (\dfrac{1}{2}) \\= 1$