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Question: The value of \[\sin {47^ \circ } + \sin {61^ \circ } - \sin {11^ \circ } - \sin {25^ \circ } = \] ...

The value of sin⁡47∘+sin⁡61∘−sin⁡11∘−sin⁡25∘=\sin {47^ \circ } + \sin {61^ \circ } - \sin {11^ \circ } - \sin {25^ \circ } =
A.sin⁡36∘\sin {36^ \circ }
B.cos⁡36∘\cos {36^ \circ }
C.sin⁡7∘\sin {7^ \circ }
D.cos⁡7∘\cos {7^ \circ }

Explanation

Solution

In this problem we will use the combinations and trigonometric identities. We cannot directly write the values of sin or cos function. Thus we will take values of other trigonometric functions.

Formula used:
sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A - \sin B = 2\cos \dfrac{{A + B}}{2}\sin \dfrac{{A - B}}{2}
sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A + \sin B = 2\sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}

Complete step-by-step answer:
Given that,
sin⁡47∘+sin⁡61∘−sin⁡11∘−sin⁡25∘\sin {47^ \circ } + \sin {61^ \circ } - \sin {11^ \circ } - \sin {25^ \circ }
⇒sin⁡47∘−sin⁡25∘+sin⁡61∘−sin⁡11∘\Rightarrow \sin {47^ \circ } - \sin {25^ \circ } + \sin {61^ \circ } - \sin {11^ \circ }
Using the first formula mentioned above,
⇒2cos⁡47∘+25∘2sin⁡47∘−25∘2+2cos⁡61∘+11∘2sin⁡61∘−11∘2\Rightarrow 2\cos \dfrac{{{{47}^ \circ } + {{25}^ \circ }}}{2}\sin \dfrac{{{{47}^ \circ } - {{25}^ \circ }}}{2} + 2\cos \dfrac{{{{61}^ \circ } + {{11}^ \circ }}}{2}\sin \dfrac{{{{61}^ \circ } - {{11}^ \circ }}}{2}
On adding the angles,
⇒2cos⁡72∘2sin⁡22∘2+2cos⁡72∘2sin⁡50∘2\Rightarrow 2\cos \dfrac{{{{72}^ \circ }}}{2}\sin \dfrac{{{{22}^ \circ }}}{2} + 2\cos \dfrac{{{{72}^ \circ }}}{2}\sin \dfrac{{{{50}^ \circ }}}{2}
⇒2cos⁡36∘sin⁡11∘+2cos⁡36∘sin⁡25∘\Rightarrow 2\cos {36^ \circ }\sin {11^ \circ } + 2\cos {36^ \circ }\sin {25^ \circ }
Taking the cos angle common,
⇒2cos⁡36∘(sin⁡11∘+sin⁡25∘)\Rightarrow 2\cos {36^ \circ }(\sin {11^ \circ } + \sin {25^ \circ })
On using the second identity formula,
⇒2cos⁡36∘(2sin⁡11∘+25∘2cos⁡11∘−25∘2)\Rightarrow 2\cos {36^ \circ }\left( {2\sin \dfrac{{{{11}^ \circ } + {{25}^ \circ }}}{2}\cos \dfrac{{{{11}^ \circ } - {{25}^ \circ }}}{2}} \right)
⇒2cos⁡36∘(2sin⁡36∘2cos⁡−14∘2)\Rightarrow 2\cos {36^ \circ }\left( {2\sin \dfrac{{{{36}^ \circ }}}{2}\cos \dfrac{{ - {{14}^ \circ }}}{2}} \right)
⇒2cos⁡36∘(2sin⁡18∘cos⁡(−7∘))\Rightarrow 2\cos {36^ \circ }\left( {2\sin {{18}^ \circ }\cos \left( { - {7^ \circ }} \right)} \right)
cos⁡(−θ)=cos⁡θ\cos \left( { - \theta } \right) = \cos \theta
⇒2cos⁡36∘(2sin⁡18∘cos⁡(7∘))\Rightarrow 2\cos {36^ \circ }\left( {2\sin {{18}^ \circ }\cos \left( {{7^ \circ }} \right)} \right)
⇒4cos⁡36∘sin⁡18∘cos⁡7∘\Rightarrow 4\cos {36^ \circ }\sin {18^ \circ }\cos {7^ \circ }
Putting the values of the angles as cos⁡36∘=5+14\cos {36^ \circ } = \dfrac{{\sqrt 5 + 1}}{4} and sin⁡18∘=5−14\sin {18^ \circ } = \dfrac{{\sqrt 5 - 1}}{4}
Thus,
⇒4×5+14×5−14×cos⁡7∘\Rightarrow 4 \times \dfrac{{\sqrt 5 + 1}}{4} \times \dfrac{{\sqrt 5 - 1}}{4} \times \cos {7^ \circ }
The values are nothing but (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}
⇒4×(5)2−124×4×cos⁡7∘\Rightarrow 4 \times \dfrac{{{{\left( {\sqrt 5 } \right)}^2} - {1^2}}}{{4 \times 4}} \times \cos {7^ \circ }
Taking the squares,
⇒5−14×cos⁡7∘\Rightarrow \dfrac{{5 - 1}}{4} \times \cos {7^ \circ }
⇒cos⁡7∘\Rightarrow \cos {7^ \circ }
This is the value of the given expression.
Thus option D is the correct option.

Note: We cannot directly put the values. We need to first use the identities with the help of which we can make the expression easier to solve. And at a point we used the values of the angles. Sometimes we also try to adjust the angles but here we have used the formulas only!