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Question: Find the value of \(8\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right)\left( ...

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

Explanation

Solution

Rewrite the equation to use the formula 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right) in the expression (2)(sin⁡12∘)(sin⁡48∘)\left( 2 \right)\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right). Further, apply the formula 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)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(sin⁡12∘)(sin⁡48∘)(sin⁡54∘)8\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right)\left( {\sin {{54}^ \circ }} \right)
We know that 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)
We can rewrite the equation as 4(2)(sin⁡12∘)(sin⁡48∘)(sin⁡54∘)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 (2)(sin⁡12∘)(sin⁡48∘)\left( 2 \right)\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right), we will get
(2)(sin⁡12∘)(sin⁡48∘)=cos⁡(12∘−48∘)−cos⁡(12∘+48∘) ⇒(2)(sin⁡12∘)(sin⁡48∘)=cos⁡(−36∘)−cos⁡(60∘)  \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⁡(−θ)=cos⁡θ\cos \left( { - \theta } \right) = \cos \theta and cos⁡60∘=12\cos {60^ \circ } = \dfrac{1}{2}
On substituting the values in the above equation, we get
(2)(sin⁡12∘)(sin⁡48∘)=cos⁡(36∘)−12\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 (2)(sin⁡12∘)(sin⁡48∘)\left( 2 \right)\left( {\sin {{12}^ \circ }} \right)\left( {\sin {{48}^ \circ }} \right) in equation (1)
4(2)(sin⁡12∘)(sin⁡48∘)(sin⁡54∘)=4(cos⁡36∘−12)(sin⁡54∘)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(cos⁡36∘−12)(sin⁡54∘)=(4cos⁡36∘−2)(sin⁡54∘) ⇒4cos⁡36∘sin⁡54∘−2sin⁡54∘  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⁡θ=cos⁡(90∘−θ)\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)
Therefore,
sin⁡54∘=cos⁡(90∘−54∘) =cos⁡(36∘)  \sin {54^ \circ } = \cos \left( {{{90}^ \circ } - {{54}^ \circ }} \right) \\\ = \cos \left( {{{36}^ \circ }} \right) \\\
Hence, the expression 4cos⁡36∘sin⁡54∘−2sin⁡54∘4\cos {36^ \circ }\sin {54^ \circ } - 2\sin {54^ \circ } can be written as,
4cos⁡36∘cos⁡36∘−2cos⁡36∘=4cos⁡236∘−2cos⁡36∘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
4cos⁡36∘cos⁡36∘−2cos⁡36∘=2(2cos⁡236∘−cos⁡36∘)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 2cos⁡2x=1+cos⁡2x2{\cos ^2}x = 1 + \cos 2x, we have
4cos⁡36∘cos⁡36∘−2cos⁡36∘=2(1+cos⁡72∘−cos⁡36∘)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=−2sin⁡(A+B2)sin⁡(A−B2)\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
2(1+cos⁡72∘−cos⁡36∘)=2+2(cos⁡72∘−cos⁡36∘) ⇒2+2(cos⁡72∘−cos⁡36∘)=2−4sin⁡(72∘+36∘2)sin⁡(72∘−36∘2) ⇒2+2(cos⁡72∘−cos⁡36∘)=2−4sin⁡54∘sin⁡18∘  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⁡θ=cos⁡(90∘−θ)\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)
Then sin⁡54∘=cos⁡(36∘)\sin {54^ \circ } = \cos \left( {{{36}^ \circ }} \right)
So, we have the expression as 2−4sin⁡54∘sin⁡18∘=2−4cos⁡36∘sin⁡18∘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∘\cos {18^ \circ } the expression 2−4cos⁡36∘sin⁡18∘2 - 4\cos {36^ \circ }\sin {18^ \circ }
Then, we have 2−4cos⁡36∘sin⁡18∘cos⁡18∘cos⁡18∘2 - \dfrac{{4\cos {{36}^ \circ }\sin {{18}^ \circ }\cos {{18}^ \circ }}}{{\cos {{18}^ \circ }}}
Since, 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x = \sin 2x
Then, 2−4cos⁡36∘sin⁡18∘cos⁡18∘cos⁡18∘=2−2cos⁡36∘sin⁡36∘cos⁡18∘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, 2−2cos⁡36∘sin⁡36∘cos⁡18∘=2−sin⁡72∘cos⁡18∘2 - \dfrac{{2\cos {{36}^ \circ }\sin {{36}^ \circ }}}{{\cos {{18}^ \circ }}} = 2 - \dfrac{{\sin {{72}^ \circ }}}{{\cos {{18}^ \circ }}}
Now,
sin⁡72∘=cos⁡(90∘−72∘) sin⁡72∘=cos⁡18∘  \sin {72^ \circ } = \cos \left( {{{90}^ \circ } - {{72}^ \circ }} \right) \\\ \sin {72^ \circ } = \cos {18^ \circ } \\\
On substituting the value, we get
2−cos⁡18∘cos⁡18∘=2−1 ⇒2−1=1  2 - \dfrac{{\cos {{18}^ \circ }}}{{\cos {{18}^ \circ }}} = 2 - 1 \\\ \Rightarrow 2 - 1 = 1 \\\
Hence, the value of 8(sin⁡12∘)(sin⁡48∘)(sin⁡54∘)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⁡θ=cos⁡(90∘−θ)\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)