Question

$\sin \frac{\pi }{5}\sin \frac{2\pi }{5}\sin \frac{3\pi }{5}\sin \frac{4\pi }{5}=?$

Answer 1

$\begin{array}{c}\sin \frac{\pi }{5}\sin \frac{2\pi }{5}\sin \frac{3\pi }{5}\sin \frac{4\pi }{5}=?\\ LHS=A+B=\pi A=\pi -B\\ \sin A=\sin \left(\pi -B\right)\\ =\sin B\\ \frac{\pi }{5}+\frac{4\pi }{5}=\pi \\ \sin \frac{\pi }{5}=\sin \frac{4\pi }{5}\\ similarly\\ \frac{2\pi }{5}+\frac{3\pi }{5}=\pi \\ =\sin \frac{2\pi }{5}=\sin \frac{3\pi }{5}\end{array}$

Now, put all values in equation (i)

$\begin{array}{c}=\sin \frac{\pi }{5}\cdot \sin \frac{2\pi }{5}\sin \cdot \frac{2\pi }{5}\cdot \sin \frac{\pi }{5}={\left(\sin \frac{\pi }{5}\sin \frac{2\pi }{5}\right)}^2\\ ={\left(\sin {36}^0\sin {72}^0\right)}^2\\ u\sin gproperty\\ \sin \left({90}^0-\theta \right)=\cos \theta \\ \cos \theta \left({90}^0-\theta \right)=\sin \theta \\ \Rightarrow {\left[\left(\sin {36}^0\cdot \cos \left({90}^0-{72}^0\right)\right)\right]}^2\\ \Rightarrow {\left[\sin 36\cos {18}^0\right]}^2\\ \Rightarrow \left[\frac{\sqrt{10-2\sqrt{5}}}{4}\cdot \frac{\sqrt{10+2\sqrt{5}}}{4}\right]\\ \Rightarrow \frac{\left(10-2\sqrt{5}\right)\left(10+2\sqrt{5}\right)}{16\cdot 16}=\frac{100-20}{256}\\ =\frac{80}{256}=\frac{5}{16}\end{array}$