Question

Prove that : $\sin {20}^{\circ }\sin {40}^{\circ }\sin {60}^{\circ }\sin {80}^{\circ }={\displaystyle \frac{3}{16}}$

Answer 1

Hint: Use trigonometry identity $2\sin A\sin B=\cos \left(A-B\right)-\cos \left(A+B\right)$
and $2\cos A\sin B=\sin \left(A+B\right)+\sin \left(A-B\right)$

Here we have to prove Left hand side(LHS) equal to Right hand side(RHS).
Let’s take a left hand side of the question.
  $LHS=\sin {20}^{\circ }\sin {40}^{\circ }\sin {60}^{\circ }\sin {80}^{\circ }$
As we know the value of $\sin {60}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}}$
$\Rightarrow {\displaystyle \frac{\sqrt{3}}{2}}\left(\sin {20}^{\circ }\sin {40}^{\circ }\sin {80}^{\circ }\right)$
Now, multiply by 2 in numerator and denominator
$$\begin{gathered} \Rightarrow {\displaystyle \frac{\sqrt{3}}{4}}\left(2\sin {20}^{\circ }\sin {40}^{\circ }\sin {80}^{\circ }\right) \\ \Rightarrow {\displaystyle \frac{\sqrt{3}}{4}}\left(\left(2\sin {20}^{\circ }\sin {40}^{\circ }\right)\sin {80}^{\circ }\right) \end{gathered}$$
Use identity $2\sin A\sin B=\cos \left(A-B\right)-\cos \left(A+B\right)$
$$\begin{gathered} \Rightarrow {\displaystyle \frac{\sqrt{3}}{4}}\left(\left(\cos \left({40}^{\circ }-{20}^{\circ }\right)-\cos \left({40}^{\circ }+{20}^{\circ }\right)\right)\sin {80}^{\circ }\right) \\ \Rightarrow {\displaystyle \frac{\sqrt{3}}{4}}\left(\left(\cos \left({20}^{\circ }\right)-\cos \left({60}^{\circ }\right)\right)\sin {80}^{\circ }\right) \\ \therefore \cos {60}^{\circ }={\displaystyle \frac{1}{2}} \\ \Rightarrow {\displaystyle \frac{\sqrt{3}}{4}}\left(\cos {20}^{\circ }\sin {80}^{\circ }\right)-{\displaystyle \frac{\sqrt{3}}{8}}\sin {80}^{\circ } \\ \Rightarrow {\displaystyle \frac{\sqrt{3}}{8}}\left(2\cos {20}^{\circ }\sin {80}^{\circ }\right)-{\displaystyle \frac{\sqrt{3}}{8}}\sin {80}^{\circ } \end{gathered}$$
Use identity $2\cos A\sin B=\sin \left(A+B\right)+\sin \left(A-B\right)$
 $$\begin{gathered} \Rightarrow {\displaystyle \frac{\sqrt{3}}{8}}\left(\sin {100}^{\circ }+\sin {60}^{\circ }\right)-{\displaystyle \frac{\sqrt{3}}{8}}\sin {80}^{\circ } \\ \Rightarrow {\displaystyle \frac{\sqrt{3}}{8}}\sin {60}^{\circ }+{\displaystyle \frac{\sqrt{3}}{8}}\sin {100}^{\circ }-{\displaystyle \frac{\sqrt{3}}{8}}\sin {80}^{\circ } \end{gathered}$$
As we know $\sin \left({180}^{\circ }-A\right)=\sin \left(A\right)$
$$\begin{gathered} \Rightarrow {\displaystyle \frac{\sqrt{3}}{8}}\sin {60}^{\circ }+{\displaystyle \frac{\sqrt{3}}{8}}\sin \left({180}^{\circ }-{80}^{\circ }\right)-{\displaystyle \frac{\sqrt{3}}{8}}\sin {80}^{\circ } \\ \Rightarrow {\displaystyle \frac{\sqrt{3}}{8}}\sin {60}^{\circ }+{\displaystyle \frac{\sqrt{3}}{8}}\sin \left({80}^{\circ }\right)-{\displaystyle \frac{\sqrt{3}}{8}}\sin {80}^{\circ } \\ \Rightarrow {\displaystyle \frac{\sqrt{3}}{8}}\sin {60}^{\circ } \\ \Rightarrow {\displaystyle \frac{\sqrt{3}}{8}}\times {\displaystyle \frac{\sqrt{3}}{2}} \\ \Rightarrow LHS={\displaystyle \frac{3}{16}} \end{gathered}$$
So, $LHS={\displaystyle \frac{3}{16}}=RHS$
Hence proved, $\sin {20}^{\circ }\sin {40}^{\circ }\sin {60}^{\circ }\sin {80}^{\circ }={\displaystyle \frac{3}{16}}$

Note:Whenever we come across these types of problems first substitute the values of known trigonometric angles and collect the rest of trigonometric terms then use the product to sum formulas to convert unknown trigonometric angles to known trigonometric angles .