find the rate of change $f(x) = 4\sin^3 x$ when $x = \frac{5\pi}{6}$ find the rate of change $f(x) = 4\sin^3 x$ when $x = \frac{5\pi}{6}$
To find the rate of change I need to find $\frac{dy}{dx}$ using the chain rule $h'(x) = g'(f(x)).f'(x)$
$g'(f(x)) = 12\sin^2 x$
$f'(x) = \cos x$
$h'(x) = 12\sin^2 x \cos x$
After this, I am completely stuck, this comes from a paper that does not allow calculators and am I supposed to know what $\frac{5\pi}6$ means and be able to apply it to the derivative function? If so could somebody offer a link or something for me to learn becasue I do not know this.

Trig by Reference Triangles:
The angle $\frac{5 \pi}{6}$ in radians will be given by $150^o$. This simple conversion can be done by remembering that $180^o = \pi$  $rad $.
To solve for
$h'(x)=12 sin^2x cosx$
$h'( \frac{ 5 \pi}{6} )=12 sin^2 ( \frac{5 \pi}{6}) cos( \frac{5 \pi}{6} )$
we can use right-angled triangles of the standard angles. E.g. here  our reference triangle is the $(90, 60, 30^o)$ triangle in the secondant quadrant of the cartesian plane. However, we are more interested in the radian form of the triangle ($ \frac{\pi}{2},  \frac{ \pi}{3},  \frac{\pi}{6}) $.  The respective side lengths of this triangle are therefore  $(2, - \sqrt{3}, 1)$. 
$sin (\frac{5 \pi}{6} ) = \frac{ - \sqrt {3}}{2}$
and, 
$-cos (\frac{5 \pi}{6} ) = \frac{1}{2}$
(See link:Reference Triangles)
