Question

Solve the given trigonometric expression: $\sin {36^\circ} \times \sin {72^\circ} \times \sin {108^\circ} \times \sin {144^\circ} = ?$
A) $\dfrac{3}{{16}}$
B) $\dfrac{1}{4}$
C) $\dfrac{5}{{16}}$
D) $\dfrac{1}{2}$

Answer 1

Hint: According to given in the question we have to solve the given expression $\sin {36^\circ} \times \sin {72^\circ} \times \sin {108^\circ} \times \sin {144^\circ}$ so, first of all we have to convert the trigonometric term $\sin {108^\circ}$ into $\sin {36^\circ}$ and same as we have to convert the trigonometric term $\sin {144^\circ}$ into $\sin {72^\circ}$

Formula used: $\sin ({180^\circ} - \theta ) = \sin \theta .................(1)$
So that we can solve the given trigonometric expression easily and after that to obtain the value of
${(\sin {36^\circ})^2}$ and ${(\sin {72^\circ})^2}$
Hence, to find the value of ${(\sin {36^\circ})^2}$ we have to follow the process given below:
First of all we have to the formula $\cos 2\theta = 1 - 2{\sin ^2}\theta .............(2)$ and we know that the value if $\sin {18^\circ} = \left( {\dfrac{{\sqrt 5 - 1}}{4}} \right)$ so, with the help of formula (2) and value of $\sin {18^\circ}$we can obtain the value of ${(\sin {36^\circ})^2}$
Now, same as we have to find the value of ${(\sin {36^\circ})^2}$ and to find the value first of all we have to find the value of $\cos {18^\circ}$ with the help of the value of $\sin {18^\circ}$.
Hence, after substituting the values in the obtained trigonometric expression we can simplify it.
$\sin ({90^\circ} - \theta ) = \cos \theta .............(a)$

Complete step-by-step answer:
Step 1: First of all we have to convert the trigonometric term $\sin {108^\circ}$ into $\sin {36^\circ}$ and same as we have to convert the trigonometric term $\sin {144^\circ}$ into $\sin {72^\circ}$ with the help of the formula (1) mentioned in the solution hint.
$ = \sin {36^\circ} \times \sin {72^\circ} \times \sin ({180^\circ} - {36^\circ}) \times \sin ({180^\circ} - {72^\circ})$
$
   = \sin {36^\circ} \times \sin {72^\circ} \times \sin {36^\circ} \times \sin {72^\circ} \\
   = {(\sin {36^\circ})^2} \times {(\sin {72^\circ})^2}.............(3)
 $
Step 2: Now, to solve the expression (3) obtained just above first of all we have to find the value of $\sin {36^\circ}$ and to find the value of $\sin {36^\circ}$ we have to use the value of $\sin {18^\circ}$ and the formula (2) as mentioned in the solution hint.
\[
   \Rightarrow \cos {36^\circ} = 1 - 2{\sin ^2}{18^\circ} \\
   \Rightarrow \cos {36^\circ} = 1 - 2{\left( {\dfrac{{\sqrt 5 - 1}}{4}} \right)^2} \\
   \Rightarrow \cos {36^\circ} = 1 - \left( {\dfrac{{5 + 1 - 2\sqrt 5 }}{4}} \right) \\
   \Rightarrow \cos {36^\circ} = 1 - \left( {\dfrac{{6 - 2\sqrt 5 }}{8}} \right) \\
   \Rightarrow \cos {36^\circ} = \dfrac{{8 - 6 + 2\sqrt 5 }}{8} \\
   \Rightarrow \cos {36^\circ} = \dfrac{{2 + 2\sqrt 5 }}{8}
 \]
Now, we have to take 2 as a common term from the numerator and divide it with 8 in the denominator.
\[ \Rightarrow \cos {36^\circ} = \dfrac{{\sqrt 5 + 1}}{8}\]
Step 3: Now, we have to find the value of ${(\sin {36^\circ})^2}$ with the help of the value of $\cos {36^\circ}$ as obtained in the step 2:
$
   \Rightarrow {\sin ^2}{36^\circ} + {\cos ^2}{36^\circ} = 1 \\
   \Rightarrow {\sin ^2}{36^\circ} = 1 - {\cos ^2}{36^\circ}
 $
On substituting the value of $\cos {36^\circ}$ in the expression obtained just above,
$
   \Rightarrow {\sin ^2}{36^\circ} = 1 - {\left( {\dfrac{{\sqrt 5 + 1}}{4}} \right)^2} \\
   \Rightarrow {\sin ^2}{36^\circ} = 1 - \left( {\dfrac{{5 + 1 + 2\sqrt 5 }}{{16}}} \right) \\
   \Rightarrow {\sin ^2}{36^\circ} = \dfrac{{16 - 6 - 2\sqrt 5 }}{{16}} \\
   \Rightarrow {\sin ^2}{36^\circ} = \dfrac{{10 - 2\sqrt 5 }}{{16}}
 $
Step 4: Now, we have to find the value of $\sin {72^\circ}$ and to find the value of $\sin {72^\circ}$ we have to use the value of $\sin {18^\circ}$ as mentioned in the solution hint.
\[
   \Rightarrow {\sin ^2}{18^\circ} + {\cos ^2}{18^\circ} = 1 \\
   \Rightarrow {\cos ^2}{18^\circ} = 1 - {\sin ^2}{18^\circ}
 \]
On substituting the value of $\sin {18^\circ}$ in the expression obtained just above,
$
   \Rightarrow \cos {18^\circ} = \sqrt {1 - {{\left( {\dfrac{{1 - \sqrt 5 }}{4}} \right)}^2}} \\
   \Rightarrow \cos {18^\circ} = \sqrt {1 - \left( {\dfrac{{1 + 5 - 2\sqrt 5 }}{{16}}} \right)} \\
   \Rightarrow \cos {18^\circ} = \sqrt {\dfrac{{16 - 6 + 2\sqrt 5 }}{{16}}} \\
   \Rightarrow \cos {18^\circ} = \sqrt {\dfrac{{10 + 2\sqrt 5 }}{4}}
 $
Now, to find the value of $\sin {72^\circ}$ we have to use the formula (a) as mentioned in the solution hint.
$
   \Rightarrow \sin {72^\circ} = \sin ({90^\circ} - {18^\circ}) \\
   \Rightarrow \sin {72^\circ} = \cos {18^\circ} \\
   \Rightarrow \sin {72^\circ} = \sqrt {\dfrac{{10 + 2\sqrt 5 }}{4}}
 $
Step 5: Now, we have to substitute the values of $\sin {36^\circ}$ and $\sin {72^\circ}$ as obtained from the step 2 and step 4 in the expression (3)
$ = {\left[ {\sqrt {\dfrac{{10 - 2\sqrt 5 }}{{16}}} } \right]^2} \times {\left[ {\sqrt {\dfrac{{10 + 2\sqrt 5 }}{{16}}} } \right]^2}$
On solving the obtained expression,
$
   = \left[ {\dfrac{{100 - 20}}{{16 \times 16}}} \right] \\
   = \dfrac{{80}}{{256}} \\
   = \dfrac{5}{{16}}
 $
Final solution: Hence, with the help of the formulas (a), (1) and (2) we have obtained the value of the given trigonometric expression $\sin {36^\circ} \times \sin {72^\circ} \times \sin {108^\circ} \times \sin {144^\circ} = \dfrac{5}{{16}}$

Therefore the option C is the correct answer.

Note: To convert $\sin \theta $ in the form of $\cos \theta $ we can use the $\sin ({90^\circ} - \theta ) = \cos \theta $ hence we can obtain the value of $\cos {18^\circ}$ from $\sin {18^\circ}$
With the help of the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ to find the value of ${\sin ^2}\theta $ or ${\cos ^2}\theta $