Question

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

Answer 1

Hint: Convert all the angles in the range of ${{0}^{\circ }}-{{90}^{\circ }}$. Simplify the given expression by substituting the values of $\sin {{36}^{\circ }}$ and $\sin {{72}^{\circ }}$, which are $\sin {{36}^{\circ }}=\dfrac{\sqrt{10-2\sqrt{5}}}{4}$ and $\sin {{72}^{\circ }}=\dfrac{\sqrt{10+2\sqrt{5}}}{4}$.

Complete step-by-step answer:
We have to calculate the value of $\sin {{36}^{\circ }}\sin {{72}^{\circ }}\sin {{108}^{\circ }}\sin {{144}^{\circ }}$.
We will firstly simplify the given expression by writing all the angles in the range of ${{0}^{\circ }}-{{90}^{\circ }}$. We can write ${{108}^{\circ }}={{180}^{\circ }}-{{72}^{\circ }}$ and ${{144}^{\circ }}={{180}^{\circ }}-{{36}^{\circ }}$.
Thus, we have $\sin {{36}^{\circ }}\sin {{72}^{\circ }}\sin {{108}^{\circ }}\sin {{144}^{\circ }}=\sin {{36}^{\circ }}\sin {{72}^{\circ }}\sin ({{180}^{\circ }}-{{72}^{\circ }})\sin ({{180}^{\circ }}-{{36}^{\circ }})$.
We know that $\sin \left( {{180}^{\circ }}-x \right)=\sin x$.
Thus, we have $\sin ({{180}^{\circ }}-{{72}^{\circ }})=\sin {{72}^{\circ }}$ and $\sin ({{180}^{\circ }}-{{36}^{\circ }})=\sin {{36}^{\circ }}$.
So, we have $\sin {{36}^{\circ }}\sin {{72}^{\circ }}\sin {{108}^{\circ }}\sin {{144}^{\circ }}=\sin {{36}^{\circ }}\sin {{72}^{\circ }}\sin ({{180}^{\circ }}-{{72}^{\circ }})\sin ({{180}^{\circ }}-{{36}^{\circ }})={{\sin }^{2}}{{36}^{\circ }}{{\sin }^{2}}{{72}^{\circ }}$.
We know that $\sin {{36}^{\circ }}=\dfrac{\sqrt{10-2\sqrt{5}}}{4}$ and $\sin {{72}^{\circ }}=\dfrac{\sqrt{10+2\sqrt{5}}}{4}$.
Substituting the above values, we have $\sin {{36}^{\circ }}\sin {{72}^{\circ }}\sin {{108}^{\circ }}\sin {{144}^{\circ }}={{\sin }^{2}}{{36}^{\circ }}{{\sin }^{2}}{{72}^{\circ }}={{\left( \dfrac{\sqrt{10-2\sqrt{5}}}{4} \right)}^{2}}{{\left( \dfrac{\sqrt{10+2\sqrt{5}}}{4} \right)}^{2}}$.
Simplifying the above equation, we have $\sin {{36}^{\circ }}\sin {{72}^{\circ }}\sin {{108}^{\circ }}\sin {{144}^{\circ }}={{\left( \dfrac{\sqrt{10-2\sqrt{5}}}{4} \right)}^{2}}{{\left( \dfrac{\sqrt{10+2\sqrt{5}}}{4} \right)}^{2}}=\dfrac{10-2\sqrt{5}}{16}\times \dfrac{10+2\sqrt{5}}{16}$.
We know the identity $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$.
Substituting $a=10,b=2\sqrt{5}$ in the above equation, we have $\left( 10+2\sqrt{5} \right)\left( 10-2\sqrt{5} \right)={{\left( 10 \right)}^{2}}-{{\left( 2\sqrt{5} \right)}^{2}}=100-20=80$.
Thus, we have $\sin {{36}^{\circ }}\sin {{72}^{\circ }}\sin {{108}^{\circ }}\sin {{144}^{\circ }}=\dfrac{10-2\sqrt{5}}{16}\times \dfrac{10+2\sqrt{5}}{16}=\dfrac{80}{256}=\dfrac{5}{16}$.
Hence, the value of the trigonometric expression $\sin {{36}^{\circ }}\sin {{72}^{\circ }}\sin {{108}^{\circ }}\sin {{144}^{\circ }}$ is $\dfrac{5}{16}$, which is option (d).
Trigonometric functions are real functions that relate any angle of a right-angled triangle to the ratios of any two of its sides. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius 1).

Note: To calculate the value of the given trigonometric expression, one must know the value of $\sin {{36}^{\circ }}$ and $\sin {{72}^{\circ }}$. If not, we can calculate the value of $\sin {{36}^{\circ }}$ by calculating the value of $\sin {{18}^{\circ }}$ using the fact that if $x={{18}^{\circ }}$, we have $5x={{90}^{\circ }}$. We can then use the trigonometric identity $\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y$ to calculate the value of $\sin {{36}^{\circ }}$. To calculate the value of $\sin {{72}^{\circ }}$, we have to calculate the value of $\cos {{18}^{\circ }}$ and then use the identity $\sin x=\cos \left( {{90}^{\circ }}-x \right)$. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one.