Question

Find the value of a if $\sin {{12}^{\circ }}.\sin {{48}^{\circ }}.\sin {{54}^{\circ }}=\dfrac{1}{a}$.

Answer 1

We are going to use the trigonometric identities like the transformation of sum and products of sin and cos. We also use quadrant rules for trigonometry. We try to form the identities of the same category and find the value of a.

Complete step-by-step solution
We try to form the trigonometric identity of $2\sin A.\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)$.
From the given equation of $\sin {{12}^{\circ }}.\sin {{48}^{\circ }}.\sin {{54}^{\circ }}=\dfrac{1}{a}$, we take $A={{12}^{\circ }},B={{48}^{\circ }}$.
So, the left-hand side becomes $\sin {{12}^{\circ }}.\sin {{48}^{\circ }}.\sin {{54}^{\circ }}=\dfrac{1}{2}\left( 2\sin {{12}^{\circ }}.\sin {{48}^{\circ }} \right).\sin {{54}^{\circ }}$.
We convert $2\sin {{12}^{\circ }}.\sin {{48}^{\circ }}$ as $2\sin {{12}^{\circ }}.\sin {{48}^{\circ }}=\cos \left( {{12}^{\circ }}-{{48}^{\circ }} \right)-\cos \left( {{12}^{\circ }}+{{48}^{\circ }} \right)$.
The equation becomes $2\sin {{12}^{\circ }}.\sin {{48}^{\circ }}=\cos \left( -{{36}^{\circ }} \right)-\cos \left( {{60}^{\circ }} \right)=\cos {{36}^{\circ }}-\dfrac{1}{2}$.
$\begin{aligned} & \sin {{12}^{\circ }}.\sin {{48}^{\circ }}.\sin {{54}^{\circ }} \\\ & =\dfrac{1}{2}\left( \cos {{36}^{\circ }}-\dfrac{1}{2} \right).\sin {{54}^{\circ }} \\\ & =\dfrac{1}{4}\left( 2\sin {{54}^{\circ }}.\cos {{36}^{\circ }}-\sin {{54}^{\circ }} \right) \\\ \end{aligned}$
We found a new equation where we can apply a new trigonometrical identity of $2\sin A.\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)$.
We take $A={{54}^{\circ }},B={{36}^{\circ }}$. We get $2\sin {{54}^{\circ }}.\cos {{36}^{\circ }}=\sin \left( {{54}^{\circ }}+{{36}^{\circ }} \right)+\sin \left( {{54}^{\circ }}-{{36}^{\circ }} \right)=\sin {{90}^{\circ }}+\sin {{18}^{\circ }}=1+\sin {{18}^{\circ }}$.
We also have the trigonometric identity of $\sin \left( {{90}^{\circ }}-\alpha \right)=\cos \alpha$. We apply that on $\sin {{54}^{\circ }}$ and get $\sin {{54}^{\circ }}=\sin \left( {{90}^{\circ }}-{{36}^{\circ }} \right)=\cos {{36}^{\circ }}$
So, the equation becomes

& \dfrac{1}{4}\left( 2\sin {{54}^{\circ }}.\cos {{36}^{\circ }}-\sin {{54}^{\circ }} \right) \\\ & =\dfrac{1}{4}\left( 1+\sin {{18}^{\circ }}-\sin {{54}^{\circ }} \right) \\\ & =\dfrac{1}{4}\left[ 1-\left( \sin {{54}^{\circ }}-\sin {{18}^{\circ }} \right) \right] \\\ \end{aligned}$$ Now we try to form the trigonometric identity of $$\sin A-\sin B=2\cos \dfrac{\left( A+B \right)}{2}\sin \dfrac{\left( A-B \right)}{2}$$. $$\begin{aligned} & \dfrac{1}{4}\left[ 1-\left( \sin {{54}^{\circ }}-\sin {{18}^{\circ }} \right) \right] \\\ & =\dfrac{1}{4}\left[ 1-\left( 2\cos \dfrac{\left( {{54}^{\circ }}+{{18}^{\circ }} \right)}{2}\sin \dfrac{\left( {{54}^{\circ }}-{{18}^{\circ }} \right)}{2} \right) \right] \\\ & =\dfrac{1}{4}\left[ 1-2\cos {{36}^{\circ }}\sin {{18}^{\circ }} \right] \\\ \end{aligned}$$ Now we try to multiply $$\cos {{18}^{\circ }}$$ to the term $$2\cos {{36}^{\circ }}\sin {{18}^{\circ }}$$. We have the trigonometric identity of $$2\sin A\cos A=\sin 2A$$. We get $$2\cos {{36}^{\circ }}\sin {{18}^{\circ }}\cos {{18}^{\circ }}=\cos {{36}^{\circ }}\left( 2\sin {{18}^{\circ }}\cos {{18}^{\circ }} \right)=\cos {{36}^{\circ }}\sin {{36}^{\circ }}$$. We apply the same process one more time and get $$\begin{aligned} & \cos {{36}^{\circ }}\sin {{36}^{\circ }} \\\ & =\dfrac{1}{2}\left( 2\sin {{36}^{\circ }}\cos {{36}^{\circ }} \right) \\\ & =\dfrac{1}{2}\left( \sin {{72}^{\circ }} \right) \\\ \end{aligned}$$ We again apply $$\sin \left( {{90}^{\circ }}-\alpha \right)=\cos \alpha $$ for $$\sin {{72}^{\circ }}$$ and get $$\sin {{72}^{\circ }}=\sin \left( {{90}^{\circ }}-{{18}^{\circ }} \right)=\cos {{18}^{\circ }}$$. We can’t just multiply $$\cos {{18}^{\circ }}$$ to the term $$2\cos {{36}^{\circ }}\sin {{18}^{\circ }}$$ without dividing with it also. So, the conversion of $$2\cos {{36}^{\circ }}\sin {{18}^{\circ }}$$ becomes $$2\cos {{36}^{\circ }}\sin {{18}^{\circ }}=\dfrac{2\cos {{36}^{\circ }}\sin {{18}^{\circ }}\cos {{18}^{\circ }}}{\cos {{18}^{\circ }}}=\dfrac{\dfrac{1}{2}\left( \sin {{72}^{\circ }} \right)}{\cos {{18}^{\circ }}}=\dfrac{\dfrac{1}{2}\left( \cos {{18}^{\circ }} \right)}{\cos {{18}^{\circ }}}=\dfrac{1}{2}$$. Now we replace the value in $$\dfrac{1}{4}\left[ 1-2\cos {{36}^{\circ }}\sin {{18}^{\circ }} \right]$$. $$\dfrac{1}{4}\left[ 1-2\cos {{36}^{\circ }}\sin {{18}^{\circ }} \right]=\dfrac{1}{4}\left[ 1-\dfrac{1}{2} \right]=\dfrac{1}{4}\times \dfrac{1}{2}=\dfrac{1}{8}$$. **So, the value of $\sin {{12}^{\circ }}.\sin {{48}^{\circ }}.\sin {{54}^{\circ }}$ is $\dfrac{1}{8}$. Value of a is 8.** **Note:** The solution can also be found by placing the values from the known angles. We find the value for angle ${{36}^{\circ }}$ and find the solution in much smaller steps. But the problem will be that the angle ${{36}^{\circ }}$ is not considered as the general value. So, we can’t always get away with using direct values.