Question

The value of $\sin {36^ \circ }\sin {72^ \circ }\sin {108^ \circ }\sin {144^ \circ }$is:
(a) $\dfrac{1}{4}$
(b) $\dfrac{1}{6}$
(c) $\dfrac{3}{4}$
(d) $\dfrac{5}{{16}}$

Answer 1

Trigonometric formulae that will be helpful in solving such questions:

1. $2\sin A \times \sin B = \cos (A - B) - \cos (A + B)$
2. $\sin ( - \theta ) = - \sin (\theta )$
3. $\cos ( - \theta ) = \cos (\theta )$

Complete step by step solution:
Given: $\sin {36^ \circ }\sin {72^ \circ }\sin {108^ \circ }\sin {144^ \circ }$

$$\Rightarrow \sin {36^ \circ }\sin {72^ \circ }\sin {(180 - 72)^ \circ }\sin {(180 - 36)^ \circ } \\\ \Rightarrow \sin {36^ \circ }\sin {72^ \circ }\sin {72^ \circ }\sin {36^ \circ }......\left( {\because \sin ({{180}^ \circ } - \theta ) = \sin \theta } \right) \\\ \Rightarrow {\left( {\sin {{36}^ \circ }} \right)^2}{\left( {\sin {{72}^ \circ }} \right)^2} \\\ \Rightarrow {\left( {2\sin {{72}^ \circ } \times \sin {{36}^ \circ }} \right)^2} \\\$$

Multiplying and dividing by 2 we get;

\Rightarrow \dfrac{1}{4}\left\\{ {{{\left( {2\sin {{72}^ \circ } \times \sin {{36}^ \circ }} \right)}^2}} \right\\} \\\ \Rightarrow \dfrac{1}{4}{\left( {\cos {{(72 - 36)}^ \circ } - \cos {{(72 + 36)}^ \circ }} \right)^2}......\left\\{ {Using{\text{ }}2\sin A \times \sin B = \cos (A - B) - \cos (A + B)} \right\\} \\\ \Rightarrow \dfrac{1}{4}{\left( {\cos {{36}^ \circ } - \cos {{108}^ \circ }} \right)^2} \\\ \Rightarrow \dfrac{1}{4}{\left( {\cos {{36}^ \circ } - \cos {{108}^ \circ }} \right)^2} \\\ \Rightarrow \dfrac{1}{4}{\left\\{ {\cos {{36}^ \circ } - \cos {{(90 + 18)}^ \circ }} \right\\}^2} \\\ \Rightarrow \dfrac{1}{4}{\left\\{ {\cos {{36}^ \circ } + \sin {{18}^ \circ }} \right\\}^2}....................Eq:01 \\\

Now, let us find the value of $\sin {18^ \circ }$.
Let, $\theta = {18^ \circ }$
Multiplying both sides by 5 we get;

$$\Rightarrow 5\theta = {90^ \circ } \\\ \Rightarrow 2\theta + 3\theta = {90^ \circ } \\\ \Rightarrow 2\theta = {90^ \circ } - 3\theta \\\$$

On applying $\sin \theta$both sides,

$$\Rightarrow \sin (2\theta ) = \sin ({90^ \circ } - 3\theta ) \\\ \Rightarrow \sin (2\theta ) = \operatorname{co} (3\theta )......\because \left( {\sin ({{90}^ \circ } - \theta ) = \cos \theta } \right) \\\ \Rightarrow 2\sin \theta \cos \theta = 4{\cos ^3}\theta - 3\cos \theta \\\ \Rightarrow 2\sin \theta = 4{\cos ^2}\theta - 3 \\\ \Rightarrow 2\sin \theta = 4\left( {1 - {{\sin }^2}\theta } \right) - 3 \\\ \Rightarrow 4{\sin ^2}\theta + 2\sin \theta - 1 = 0 \\\$$

Put (x) in place of $\sin \theta$,
$\Rightarrow 4{x^2} + 2x - 1 = 0$
Using discriminant formula, solve the above quadratic equation;

$$\Rightarrow x = \dfrac{{ - 2 \pm \sqrt {{{(2)}^2} - 4(4)( - 1)} }}{{2(4)}} \\\ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 + 16} }}{{2(4)}} \\\ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {20} }}{{2(4)}} \\\ \Rightarrow x = \dfrac{{ - 2 \pm 2\sqrt 5 }}{{2(4)}} \\\ \Rightarrow x = \dfrac{{ \pm \sqrt 5 - 1}}{4} \\\$$

Putting back the value of x and $\theta$in above equation,
$\Rightarrow \sin {18^ \circ } = \dfrac{{\sqrt 5 - 1}}{4}......\because \left[ {{{18}^ \circ }\;lie{\text{ }}in{\text{ }}first{\text{ }}quadrant{\text{ }}so{\text{ }}it{\text{ }}must{\text{ }}be{\text{ }}positive} \right]$
Similarly, the value of $\cos {36^ \circ } = \dfrac{{\sqrt 5 + 1}}{4}$.
Putting back the values of $\sin {18^ \circ }$and $\cos {36^ \circ }$in Eq:01,

\Rightarrow \dfrac{1}{4}{\left\\{ {\dfrac{{\sqrt 5 + 1}}{4} + \dfrac{{\sqrt 5 - 1}}{4}} \right\\}^2} \\\ \Rightarrow \dfrac{1}{4}{\left\\{ {2\dfrac{{\sqrt 5 }}{4}} \right\\}^2} \\\ \Rightarrow \dfrac{1}{4}{\left\\{ {\dfrac{{\sqrt 5 }}{2}} \right\\}^2} \\\ \Rightarrow \dfrac{1}{4}\left( {\dfrac{5}{4}} \right) = \dfrac{5}{{16}} \\\

Our required value is $\dfrac{5}{{16}}$.

Option (D) is correct.

Note:
Quadrant plays an important role in trigonometric questions. In quadrant 1st all trigonometric ratios are positive, in 2nd quadrant only sin and cosec trigonometric ratios are positive rest negative, in 3rd quadrant only tan and cot trigonometric ratios are positive test negative and in 4th quadrant only cos and sec trigonometric ratios are positive rest negative.