Is there a simple way to calculate $\sin \frac{3\pi}{10}-\sin \frac{\pi}{10}$?

Question

I know how to find the exact value of $\sin \frac{\pi}{10}$ using double and triple angle formulas and the fact that $\frac{5\pi}{10}=\frac{\pi}{2}$ but it maybe too complicated for high school students. Is there an easier way that I do not see? The answer is $0.5$.

Unfortunately I did not see a clear answer among posted answers to my question but thanks to @labbhattacharjee, the main idea is to multiply and divide by $2 \cos 18°$.
Thus $\large{\sin 54°-\sin 18°=\frac{2 \sin 54°\cos 18°-2 \sin 18°\cos 18°}{2 \cos 18°}=\frac{\sin 72° + \sin 36°-\sin 36°}{2 \cos 18°}=\frac{\sin 72°}{2\cos 18°}=0.5}$

Answer 1

Observe \begin{align} \sin x-\sin y=2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right) \end{align} then it follows \begin{align} \sin \frac{3\pi}{10}-\sin\frac{\pi}{10} = 2\sin\frac{\pi}{10}\cos\frac{2\pi}{10} = 2\sin\frac{\pi}{10}\left(\cos^2\frac{\pi}{10}-\sin^2\frac{\pi}{10} \right). \end{align}

Answer 2

\begin{align} \sin \frac{3\pi}{10}&=\cos(\frac{\pi}{2} - \frac{3\pi}{10})\\ &\Rightarrow \cos(\frac{\pi}{2} - \frac{3\pi}{10})-\sin\frac{\pi}{10} \\ &=\cos\frac{\pi}{5} -\sin\frac{\pi}{10} \end{align}

Answer 3

That is $$\cos\frac{\pi}{5}-\cos\frac{2\pi}5=-\cos\frac{4\pi}{5}-\cos\frac{2\pi}5=A$$ say. Then $$A\sin\frac{\pi}5=-\sin\frac{\pi}5\left(\cos\frac{4\pi}{5}+\cos\frac{2\pi}5\right)=-\frac12\left(\sin\frac{5\pi}5-\sin\frac{3\pi}5+\sin\frac{3\pi}5- \sin\frac{\pi}5\right)=\frac12\sin\frac\pi5$$ etc.