Question

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

Answer 1

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) \\\ \end{aligned}$$ The expression $$\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)$$ $$\begin{aligned} & \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} \\\ \end{aligned}$$ Substituting this value in equation (1), we get $$\begin{aligned} & \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{8}=\dfrac{k}{16} \\\ & \Rightarrow \dfrac{3}{16}=\dfrac{k}{16} \\\ \end{aligned}$$ Multiplying both sides of this equation by 16, we get $3=k$ Thus the value of k is 3. Note: The formula $$\sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)=\dfrac{1}{4}\sin 3\theta $$ should be used carefully and works only for those values of $\theta $ for which the value of $\sin 3\theta $ is known to us. Derivation of the formula can be found by using the formula $\sin A\sin B=\dfrac{1}{2}\left( \cos \left( A-B \right)-\cos \left( A+B \right) \right)$ for $A={{60}^{\circ }}-\theta $ and $B={{60}^{\circ }}+\theta $. Thus, the expression becomes $$\begin{aligned} & \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) \\\ \end{aligned}$$ Now, we know that $\cos \left( -\theta \right)=\cos \theta $ and $\cos {{120}^{\circ }}=\dfrac{-1}{2}$. Using these values in the above equation, we get $$\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)$$ In this equation, substitute $\cos 2\theta =1-2{{\sin }^{2}}\theta $. $$\begin{aligned} & \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) \\\ \end{aligned}$$ Now we know that $$\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)$$.