Question

Prove that $\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}=\dfrac{3}{16}$ .

Answer 1

Hint : We first try to break the multiplication into two parts. We find the multiplication of $\sin {{40}^{\circ }}\sin {{80}^{\circ }}$ with the help of $2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)$ . Then we complete the other multiplication part to get to the solution.

Complete step-by-step answer :
We first try to put the values for $\sin {{60}^{\circ }}$ in $\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}$ .
We know that $\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$.
Therefore, $\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}=\left( \dfrac{\sqrt{3}}{2}\sin {{20}^{\circ }} \right)\left( \sin {{40}^{\circ }}\sin {{80}^{\circ }} \right)$
We have multiplication of three trigonometric ratios. We multiply them separately.
We have the identity theorem where $2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)$ .
For our given multiplication of $\left( \sin {{40}^{\circ }}\sin {{80}^{\circ }} \right)$, we first take $\dfrac{1}{2}\left( 2\sin {{40}^{\circ }}\sin {{80}^{\circ }} \right)$.
We take $\dfrac{1}{2}\left( 2\sin {{40}^{\circ }}\sin {{80}^{\circ }} \right)=\cos \left( {{80}^{\circ }}-{{40}^{\circ }} \right)-\cos \left( {{80}^{\circ }}+{{40}^{\circ }} \right)=\cos {{40}^{\circ }}-\cos {{120}^{\circ }}$.
Now we have to find the value of $\cos {{120}^{\circ }}$ .
For general form of $\cos \left( x \right)$ , we need to convert the value of x into the closest multiple of $\dfrac{\pi }{2}$ and add or subtract a certain value $\alpha$ from that multiple of $\dfrac{\pi }{2}$ to make it equal to x.
Let’s assume $x=k\times \dfrac{\pi }{2}+\alpha$ , $k\in \mathbb{Z}$ . Here we took addition of $\alpha$ . We also need to remember that $\left| \alpha \right|\le \dfrac{\pi }{2}$ .
Now we take the value of k. If it’s even then keep the ratio as cos and if it’s odd then the ratio changes to sin ratio from cos.
Then we find the position of the given angle as quadrant value measured in counter clockwise movement from the origin and the positive side of X-axis.
If the angle falls in the first or fourth quadrant then the sign remains positive but if it falls in the second or third quadrant then the sign becomes negative.
The final form becomes $\cos {{120}^{\circ }}=\cos \left( 1\times \dfrac{\pi }{2}+30 \right)=-\sin \left( 30 \right)=-\dfrac{1}{2}$ .
So, $\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}=\left( \dfrac{\sqrt{3}}{4}\sin {{20}^{\circ }} \right)\left( \cos {{40}^{\circ }}+\dfrac{1}{2} \right)$.
Again, $\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}=\dfrac{\sqrt{3}}{4}\left( \sin {{20}^{\circ }}\cos {{40}^{\circ }}+\dfrac{1}{2}\sin {{20}^{\circ }} \right)$.
So, $\sin {{20}^{\circ }}\cos {{40}^{\circ }}=\dfrac{1}{2}\left( 2\sin {{20}^{\circ }}\cos {{40}^{\circ }} \right)=\dfrac{1}{2}\left( \sin {{60}^{\circ }}-\sin {{20}^{\circ }} \right)=\dfrac{1}{2}\left( \dfrac{\sqrt{3}}{2}-\sin {{20}^{\circ }} \right)$
We have

& \sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }} \\\ & =\dfrac{\sqrt{3}}{4}\left( \sin {{20}^{\circ }}\cos {{40}^{\circ }}+\dfrac{1}{2}\sin {{20}^{\circ }} \right) \\\ & =\dfrac{\sqrt{3}}{4}\left( \dfrac{\sqrt{3}}{4}-\dfrac{1}{2}\sin {{20}^{\circ }}+\dfrac{1}{2}\sin {{20}^{\circ }} \right) \\\ & =\dfrac{3}{16} \\\ \end{aligned}$$. Thus proved, $ \sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}=\dfrac{3}{16} $ . **Note** : We also used the formula of $ 2\cos A\sin B=\sin \left( A+B \right)-\sin \left( A-B \right) $ for $$\sin {{20}^{\circ }}\cos {{40}^{\circ }}$$. We have to be careful about the choosing of similar trigonometric ratios.