Question

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

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.

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,
\begin{align} & \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{align}

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{align} & \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{align}

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

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{align} & \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{align}

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{align} & \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{align}

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 $\sin \theta \sin \left( {{60}^{\circ }}-\theta \right)\sin \left( {{60}^{\circ }}+\theta \right)$.