Question

Find the value of $8\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right)\left( {\sin {{54}^ \circ }} \right)$

Answer 1

Hint: Rewrite the equation to use the formula $2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)$ in the expression $\left( 2 \right)\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right)$. Further, apply the formula $2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)$ and simplify the given expression. Then, use various trigonometric identities to solve it.

Complete step by step Answer:

We have to calculate the value of $8\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right)\left( {\sin {{54}^ \circ }} \right)$
We know that $2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)$
We can rewrite the equation as $4\left( 2 \right)\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right)\left( {\sin {{54}^ \circ }} \right)$ (1)
Now, applying the formula in $\left( 2 \right)\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right)$, we will get
$
  \left( 2 \right)\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right) = \cos \left( {{{12}^ \circ } - {{48}^ \circ }} \right) - \cos \left( {{{12}^ \circ } + {{48}^ \circ }} \right) \\
   \Rightarrow \left( 2 \right)\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right) = \cos \left( { - {{36}^ \circ }} \right) - \cos \left( {{{60}^ \circ }} \right) \\
$
We know that $\cos \left( { - \theta } \right) = \cos \theta $ and \[\cos {60^ \circ } = \dfrac{1}{2}\]
On substituting the values in the above equation, we get
\[\left( 2 \right)\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right) = \cos \left( {{{36}^ \circ }} \right) - \dfrac{1}{2}\]
Substitute the value of the expression $\left( 2 \right)\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right)$ in equation (1)
$4\left( 2 \right)\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right)\left( {\sin {{54}^ \circ }} \right) = 4\left( {\cos {{36}^ \circ } - \dfrac{1}{2}} \right)\left( {\sin {{54}^ \circ }} \right)$
Now, we will simplify the expression by opening the brackets
$
  4\left( {\cos {{36}^ \circ } - \dfrac{1}{2}} \right)\left( {\sin {{54}^ \circ }} \right) = \left( {4\cos {{36}^ \circ } - 2} \right)\left( {\sin {{54}^ \circ }} \right) \\
   \Rightarrow 4\cos {36^ \circ }\sin {54^ \circ } - 2\sin {54^ \circ } \\
$
Now, we know that $\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)$
Therefore,
$
  \sin {54^ \circ } = \cos \left( {{{90}^ \circ } - {{54}^ \circ }} \right) \\
   = \cos \left( {{{36}^ \circ }} \right) \\
 $
Hence, the expression $4\cos {36^ \circ }\sin {54^ \circ } - 2\sin {54^ \circ }$ can be written as,
$4\cos {36^ \circ }\cos {36^ \circ } - 2\cos {36^ \circ } = 4{\cos ^2}{36^ \circ } - 2\cos {36^ \circ }$
Now, we will simplify the above equation by taking common term 2 out
$4\cos {36^ \circ }\cos {36^ \circ } - 2\cos {36^ \circ } = 2\left( {2{{\cos }^2}{{36}^ \circ } - \cos {{36}^ \circ }} \right)$
On applying the formula $2{\cos ^2}x = 1 + \cos 2x$, we have
$4\cos {36^ \circ }\cos {36^ \circ } - 2\cos {36^ \circ } = 2\left( {1 + \cos {{72}^ \circ } - \cos {{36}^ \circ }} \right)$
We know that $\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
$
  2\left( {1 + \cos {{72}^ \circ } - \cos {{36}^ \circ }} \right) = 2 + 2\left( {\cos {{72}^ \circ } - \cos {{36}^ \circ }} \right) \\
   \Rightarrow 2 + 2\left( {\cos {{72}^ \circ } - \cos {{36}^ \circ }} \right) = 2 - 4\sin \left( {\dfrac{{{{72}^ \circ } + {{36}^ \circ }}}{2}} \right)\sin \left( {\dfrac{{{{72}^ \circ } - {{36}^ \circ }}}{2}} \right) \\
   \Rightarrow 2 + 2\left( {\cos {{72}^ \circ } - \cos {{36}^ \circ }} \right) = 2 - 4\sin {54^ \circ }\sin {18^ \circ } \\
$
Again, $\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)$
Then $\sin {54^ \circ } = \cos \left( {{{36}^ \circ }} \right)$
So, we have the expression as $2 - 4\sin {54^ \circ }\sin {18^ \circ } = 2 - 4\cos {36^ \circ }\sin {18^ \circ }$
We know that 36 is twice of 18
Multiply and divide the expression by $\cos {18^ \circ }$ the expression $2 - 4\cos {36^ \circ }\sin {18^ \circ }$
Then, we have $2 - \dfrac{{4\cos {{36}^ \circ }\sin {{18}^ \circ }\cos {{18}^ \circ }}}{{\cos {{18}^ \circ }}}$
Since, $2\sin x\cos x = \sin 2x$
Then, $2 - \dfrac{{4\cos {{36}^ \circ }\sin {{18}^ \circ }\cos {{18}^ \circ }}}{{\cos {{18}^ \circ }}} = 2 - \dfrac{{2\cos {{36}^ \circ }\sin {{36}^ \circ }}}{{\cos {{18}^ \circ }}}$
Similarly, $2 - \dfrac{{2\cos {{36}^ \circ }\sin {{36}^ \circ }}}{{\cos {{18}^ \circ }}} = 2 - \dfrac{{\sin {{72}^ \circ }}}{{\cos {{18}^ \circ }}}$
Now,
$
  \sin {72^ \circ } = \cos \left( {{{90}^ \circ } - {{72}^ \circ }} \right) \\
  \sin {72^ \circ } = \cos {18^ \circ } \\
$
On substituting the value, we get
$
  2 - \dfrac{{\cos {{18}^ \circ }}}{{\cos {{18}^ \circ }}} = 2 - 1 \\
   \Rightarrow 2 - 1 = 1 \\
$
Hence, the value of $8\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right)\left( {\sin {{54}^ \circ }} \right)$ is 1

Note: Students must know the trigonometric identities to solve this question. Avoid calculation mistakes. Simplification of the terms is an important step in these questions. Many students get stuck by not using simplification formulas like $\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)$