Question

Prove the following trigonometric equation \[16\sin {{10}^{\circ }}\sin {{30}^{\circ }}\sin {{50}^{\circ }}\sin {{70}^{\circ }}=1\]

Answer 1

Hint: First of all we will change the sine ratios to cosine ratios by using the identity \[sin{{\left( 90-\theta \right)}^{\circ }}=\cos \theta \] and then we will use the formula of a trigonometry which is as follows:
\[2\sin A\cos A=\sin 2A\]

Complete step-by-step answer:
We have been asked to prove \[16\sin {{10}^{\circ }}\sin {{30}^{\circ }}\sin {{50}^{\circ }}\sin {{70}^{\circ }}=1\].
Taking left hand side into consideration, we have,
\[16\sin {{10}^{\circ }}\sin {{30}^{\circ }}\sin {{50}^{\circ }}\sin {{70}^{\circ }}\]
Now we can write \[\sin {{10}^{\circ }}=\sin \left( {{90}^{\circ }}-{{80}^{\circ }} \right)\] and we know that \[\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta \] .
\[\Rightarrow \sin {{10}^{\circ }}=\sin \left( {{90}^{\circ }}-{{80}^{\circ }} \right)=\cos {{80}^{\circ }}\]
Similarly,
\[\begin{align}
  & \sin {{50}^{\circ }}=\sin \left( {{90}^{\circ }}-{{40}^{\circ }} \right)=\cos {{40}^{\circ }} \\
 & \sin {{70}^{\circ }}=\sin \left( {{90}^{\circ }}-{{20}^{\circ }} \right)=\cos {{20}^{\circ }} \\
\end{align}\]
Substituting these values we get as follows:
\[16\sin {{10}^{\circ }}\sin {{30}^{\circ }}\sin {{50}^{\circ }}\sin {{70}^{\circ }}=16\cos {{80}^{\circ }}\sin {{30}^{\circ }}\cos {{40}^{\circ }}\cos {{20}^{\circ }}\]
Since we know that \[\sin {{30}^{\circ }}=\dfrac{1}{2}\], so by substituting this value we get as follows:
\[\begin{align}
  & 16\sin {{10}^{\circ }}\sin {{30}^{\circ }}\sin {{50}^{\circ }}\sin {{70}^{\circ }}=16\cos {{80}^{\circ }}\times \left( \dfrac{1}{2} \right)\times \cos {{40}^{\circ }}\cos {{20}^{\circ }} \\
 & =8\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{80}^{\circ }} \\
\end{align}\]
On multiplying and dividing by \[\left( 2\sin {{20}^{\circ }} \right)\], we get as follows:
\[=\dfrac{8\left( 2\sin {{20}^{\circ }}\cos {{20}^{\circ }} \right)\cos {{40}^{\circ }}\cos {{80}^{\circ }}}{2\sin {{20}^{\circ }}}\]
Since we know the formula \[2\sin \theta \cos \theta =\sin 2\theta \]
So by using this, we get as follows:
\[=\dfrac{4\sin {{40}^{\circ }}\cos {{40}^{\circ }}\cos {{80}^{\circ }}}{\sin {{20}^{\circ }}}=\dfrac{2\left( 2\sin {{40}^{\circ }}\cos {{40}^{\circ }} \right)\cos {{80}^{\circ }}}{\sin {{20}^{\circ }}}\]
Again, using \[2\sin \theta \cos \theta =\sin 2\theta \], we get as follows:
\[=\dfrac{2\sin {{80}^{\circ }}\cos {{80}^{\circ }}}{\sin {{20}^{\circ }}}\]
Again, using the formula \[2\sin \theta \cos \theta =\sin 2\theta \], we get as follows:
\[=\dfrac{\sin {{160}^{\circ }}}{\sin {{20}^{\circ }}}\]
We can write \[\sin {{160}^{\circ }}=\sin \left( {{180}^{\circ }}-{{20}^{\circ }} \right)\] and we know that \[\sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta \]
\[\Rightarrow \sin {{160}^{\circ }}=\sin {{20}^{\circ }}\]
So by substituting the value of \[\sin {{160}^{\circ }}\] we get as follows:
\[16\sin {{10}^{\circ }}\sin {{30}^{\circ }}\sin {{50}^{\circ }}\sin {{70}^{\circ }}=\dfrac{\sin {{20}^{\circ }}}{\sin {{20}^{\circ }}}=1\] = right hand side
Hence the given expression is proved.

Note: Be careful while using the trigonometric formula and its identity and care of the sign while doing calculation in each step. Also remember the trigonometric formula and identity which helps you a lot in these types of questions. Whenever we have this type of question, we must first try to convert it into terms such that we can apply the identities that we know. First, we changed the sine terms of non-standard angles to cosine terms. Then, here we have used the double angle identity of sine function to simplify the expression.