Answer
$0.342$
Work Step by Step
The Taylor approximation for a function $\sin x$ centred at $a=0$ can be written as:
$p(x)=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-.........$----
Re-write the expression as: $\sqrt {\sin 20^{\circ}}=\sin (20 \times \dfrac{\pi}{180})=\sin (\dfrac{\pi}{9})$
Now, we have:
$p(\dfrac{\pi}{9})=\dfrac{\pi}{9}-\dfrac{(\dfrac{\pi}{9})^3}{3!}+\dfrac{(\dfrac{\pi}{9})^5}{5!}-.......=0.342$
Therefore, $\sqrt {\sin 20^{\circ}}=0.342$
