Question

Prove the given trigonometric expression:
$\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}=\dfrac{3}{16}$

Answer 1

At first convert the value of $\sin {{60}^{\circ }}$ as $\dfrac{\sqrt{3}}{2}$ as it is a value of the standard angle. Then use the product sum rules which are $2\sin A\sin B=\cos \left( B-A \right)-\cos \left( A+B \right)$and $2\cos A\sin B=\sin \left( A+B \right)-\sin \left( A-B \right)$ Then after all this, finally use the identity $\sin \left( \pi -\theta \right)=\sin \theta$ to get the desired results.

Complete step by step answer:
We are given the expression $\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}$ and we have to prove its product is $\dfrac{3}{16}$. So at first, we will write the expression as it is
$\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}$
We can further rearrange and write it as,
$\sin {{60}^{\circ }}\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{20}^{\circ }}$
Now, by the use of standard values of the standard angles, $\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$, we can write the expression as,
$\dfrac{\sqrt{3}}{2}\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{80}^{\circ }}$
Now we will multiply and divide 2 and write it as,
$\dfrac{\sqrt{3}}{4}\left( 2\sin {{20}^{\circ }}\sin {{40}^{\circ }} \right)\sin {{80}^{\circ }}$
Now, we will apply the identity, 2 sin A sin B = cos (B – A) – cos (B + A).
$2\sin {{20}^{\circ }}\sin {{40}^{\circ }}=\cos \left( {{40}^{\circ }}-{{20}^{\circ }} \right)-\cos \left( {{40}^{\circ }}+{{20}^{\circ }} \right)$
So, the expression holds as,
$\dfrac{\sqrt{3}}{4}\left[ \cos {{20}^{\circ }}-\cos {{60}^{\circ }} \right]\sin {{80}^{\circ }}$
We know that value of $\cos {{60}^{\circ }}\text{ is }\dfrac{1}{2}$. So, we can write it as,
$\dfrac{\sqrt{3}}{4}\left( \cos {{20}^{\circ }}-\dfrac{1}{2} \right)\sin {{80}^{\circ }}$
Now, let’s expand.
$\dfrac{\sqrt{3}}{4}\cos {{20}^{\circ }}\sin {{80}^{\circ }}-\dfrac{\sqrt{3}}{8}\sin {{80}^{\circ }}$
Now, let’s rewrite the expression as
$\dfrac{\sqrt{3}}{8}\left( 2\cos {{20}^{\circ }}\sin {{80}^{\circ }} \right)-\dfrac{\sqrt{3}}{8}\sin {{80}^{\circ }}$
Here, we will apply the identity, 2 cos A sin B = sin (A + B) – sin (A – B).
$2\cos {{20}^{\circ }}\sin {{80}^{\circ }}=\sin \left( {{20}^{\circ }}+{{80}^{\circ }} \right)-\sin \left( {{20}^{\circ }}-{{80}^{\circ }} \right)$
$2\cos {{20}^{\circ }}\sin {{80}^{\circ }}=\sin \left( {{100}^{\circ }} \right)-\sin \left( -{{60}^{\circ }} \right)$
We know that the value of $\sin \left( -{{60}^{\circ }} \right)\text{ as }-\sin {{60}^{\circ }}$ which is equal to $\dfrac{-\sqrt{3}}{2}$.
Hence, we get the expression as,
$\dfrac{\sqrt{3}}{8}\left( \sin {{100}^{\circ }}-\dfrac{-\sqrt{3}}{2} \right)-\dfrac{\sqrt{3}}{8}\sin {{80}^{\circ }}$
$\dfrac{\sqrt{3}}{8}\sin {{100}^{\circ }}+\dfrac{\sqrt{3}}{8}\times \dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{8}\sin {{80}^{\circ }}$
We will apply $\sin \left( \pi -\theta \right)=\sin \theta$
$\sin \left( {{180}^{\circ }}-80 \right)=\sin {{80}^{\circ }}$
$\sin {{100}^{\circ }}=\sin {{80}^{\circ }}$
Here, we get the expression as,
$\dfrac{\sqrt{3}}{8}\left( \sin {{100}^{\circ }}-\sin {{80}^{\circ }} \right)+\dfrac{3}{16}$
$=\dfrac{3}{16}$

Note:
One can also do the same question by another method. Instead of using the identities one can check out trigonometric table to see the exact values of $\sin {{20}^{o}},\sin {{40}^{o}},\sin {{60}^{o}},\sin {{80}^{o}}$ and then find their product to get the answer. The process would be very time consuming and will not give accurate answers.