Questions & Answers

Find the value of k, if $\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}=\dfrac{k}{16}$.

Answer Verified Verified
Hint: Write the given expression on LHS as \[\sin {{60}^{\circ }}\sin {{20}^{\circ }}\sin \left( {{60}^{\circ }}-{{20}^{\circ }} \right)\sin \left( {{60}^{\circ }}+{{20}^{\circ }} \right)\]. Use the value of $\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$ and then use the formula \[\sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)=\dfrac{1}{4}\sin 3\theta \] with $\theta ={{20}^{\circ }}$. This gives the value of the expression on LHS, and a linear equation in k. Solve the linear equation to find the value of k.

Complete step-by-step answer:
We have been given $\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{60}^{\circ }}\sin {{80}^{\circ }}$on the LHS. We know the value of $\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$. Using this value, the expression becomes,

\[\dfrac{\sqrt{3}}{2}\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{80}^{\circ }}=\dfrac{k}{16}\]
Express $\sin {{40}^{\circ }}$ as $\sin \left( {{60}^{\circ }}-{{20}^{\circ }} \right)$ and $\sin {{80}^{\circ }}$ as $\sin \left( {{60}^{\circ }}+{{20}^{\circ }} \right)$. Thus, the expression becomes,
 & \dfrac{\sqrt{3}}{2}\sin {{20}^{\circ }}\sin {{40}^{\circ }}\sin {{80}^{\circ }}=\dfrac{k}{16} \\
 & \Rightarrow \dfrac{\sqrt{3}}{2}\sin {{20}^{\circ }}\sin \left( {{60}^{\circ }}-{{20}^{\circ }} \right)\sin \left( {{60}^{\circ }}+{{20}^{\circ }} \right)=\dfrac{k}{16}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 1 \right) \\

The expression \[\sin {{20}^{\circ }}\sin \left( {{60}^{\circ }}-{{20}^{\circ }} \right)\sin \left( {{60}^{\circ }}+{{20}^{\circ }} \right)\] can be written as \[\sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)\] with $\theta ={{20}^{\circ }}$.

We know that the expression \[\sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)\] can be directly found out using the formula \[\sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)=\dfrac{1}{4}\sin 3\theta \].

Thus, applying this formula to the given expression and substituting the value of $\theta ={{20}^{\circ }}$, we get
\[\sin {{20}^{\circ }}\sin \left( {{60}^{\circ }}-{{20}^{\circ }} \right)\sin \left( {{60}^{\circ }}+{{20}^{\circ }} \right)=\dfrac{1}{4}\sin \left( 3\times {{20}^{\circ }} \right)\]
 & \Rightarrow \sin {{20}^{\circ }}\sin \left( {{60}^{\circ }}-{{20}^{\circ }} \right)\sin \left( {{60}^{\circ }}+{{20}^{\circ }} \right)=\dfrac{1}{4}\sin {{60}^{\circ }} \\
 & \Rightarrow \sin {{20}^{\circ }}\sin \left( {{60}^{\circ }}-{{20}^{\circ }} \right)\sin \left( {{60}^{\circ }}+{{20}^{\circ }} \right)=\dfrac{1}{4}\times \dfrac{\sqrt{3}}{2} \\
 & \Rightarrow \sin {{20}^{\circ }}\sin \left( {{60}^{\circ }}-{{20}^{\circ }} \right)\sin \left( {{60}^{\circ }}+{{20}^{\circ }} \right)=\dfrac{\sqrt{3}}{8} \\

Substituting this value in equation (1), we get
 & \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{8}=\dfrac{k}{16} \\
 & \Rightarrow \dfrac{3}{16}=\dfrac{k}{16} \\

Multiplying both sides of this equation by 16, we get
Thus the value of k is 3.

Note: The formula \[\sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)=\dfrac{1}{4}\sin 3\theta \] should be used carefully and works only for those values of $\theta $ for which the value of $\sin 3\theta $ is known to us. Derivation of the formula can be found by using the formula $\sin A\sin B=\dfrac{1}{2}\left( \cos \left( A-B \right)-\cos \left( A+B \right) \right)$ for $A={{60}^{\circ }}-\theta $ and $B={{60}^{\circ }}+\theta $.

Thus, the expression becomes
 & \sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)=\left( \sin \theta \right)\cdot \left( \dfrac{1}{2}\left( \cos \left( {{60}^{\circ }}-\theta -\left( {{60}^{\circ }}+\theta \right) \right)-\cos \left( {{60}^{\circ }}-\theta +{{60}^{\circ }}+\theta \right) \right) \right) \\
 & \Rightarrow \sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)=\left( \sin \theta \right)\cdot \left( \dfrac{1}{2}\left( \cos \left( -2\theta \right)-\cos \left( {{120}^{\circ }} \right) \right) \right) \\

Now, we know that $\cos \left( -\theta \right)=\cos \theta $ and $\cos {{120}^{\circ }}=\dfrac{-1}{2}$. Using these values in the above equation, we get
\[\sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)=\left( \sin \theta \right)\cdot \left( \dfrac{1}{2}\left( \cos \left( 2\theta \right)+\dfrac{1}{2} \right) \right)\]
In this equation, substitute $\cos 2\theta =1-2{{\sin }^{2}}\theta $.
 & \sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)=\left( \sin \theta \right)\cdot \left( \dfrac{1}{2}\left( 1-2{{\sin }^{2}}\theta +\dfrac{1}{2} \right) \right) \\
 & \Rightarrow \sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)=\left( \sin \theta \right)\cdot \left( \dfrac{1}{2}\left( \dfrac{3}{2}-2{{\sin }^{2}}\theta \right) \right) \\
 & \Rightarrow \sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)=\left( \sin \theta \right)\cdot \left( \dfrac{3}{4}-{{\sin }^{2}}\theta \right) \\
 & \Rightarrow \sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)=\dfrac{3}{4}\sin \theta -{{\sin }^{3}}\theta \\
 & \Rightarrow \sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)=\dfrac{1}{4}\left( \sin \theta -4{{\sin }^{3}}\theta \right) \\

Now we know that \[\sin \theta -4{{\sin }^{3}}\theta =\sin 3\theta \]. Thus, the above expression becomes
\[\sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)=\dfrac{1}{4}\sin 3\theta \]
This is the required proof. It is advisable to memorize this result as it is very helpful in solving questions where the expression can be reduced to the form of \[\sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)\].
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