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Question: Choose the correct answer the value of \(\sin {47^ \circ } + \sin {61^ \circ } - \sin {11^ \circ } -...

Choose the correct answer the value of sin⁡47∘+sin⁡61∘−sin⁡11∘−sin⁡25∘\sin {47^ \circ } + \sin {61^ \circ } - \sin {11^ \circ } - \sin {25^ \circ } is
(A) cos⁡7∘\cos {7^ \circ }
(B) sin⁡14∘\sin {14^ \circ }
(C) sin⁡7∘\sin {7^ \circ }
(D) cos⁡14∘\cos {14^ \circ }

Explanation

Solution

According to the question we have to find the value of the following equation sin⁡47∘+sin⁡61∘−sin⁡11∘−sin⁡25∘\sin {47^ \circ } + \sin {61^ \circ } - \sin {11^ \circ } - \sin {25^ \circ }. To determine the value of the equation we use the formula sin⁡c−sin⁡d=2cos⁡c+d2.sin⁡c−d2\sin c - \sin d = 2\cos \dfrac{{c + d}}{2} . \sin \dfrac{{c - d}}{2}. Then put values in place of c and d we get the equation (sin⁡47∘−sin⁡25∘)+(sin⁡61∘−sin⁡11∘)=2cos⁡47+252.sin⁡47−252+2cos⁡61+112.sin⁡61−112(\sin {47^ \circ } - \sin {25^ \circ }) + (\sin {61^ \circ } - \sin {11^ \circ }) = 2\cos \dfrac{{47 + 25}}{2} . \sin \dfrac{{47 - 25}}{2} + 2\cos \dfrac{{61 + 11}}{2} . \sin \dfrac{{61 - 11}}{2}.By solving this we get the value of the equation.

Complete step-by-step answer:
In the question we will find the value of sin⁡47∘+sin⁡61∘−sin⁡11∘−sin⁡25∘\sin {47^ \circ } + \sin {61^ \circ } - \sin {11^ \circ } - \sin {25^ \circ }
First we arrange the equation in the following form sin⁡47∘−sin⁡25∘+sin⁡61∘−sin⁡11∘\sin {47^ \circ } - \sin {25^ \circ } + \sin {61^ \circ } - \sin {11^ \circ }
For finding the value we use the formula
→sin⁡c−sin⁡d=2cos⁡c+d2.sin⁡c−d2\to \sin c - \sin d = 2\cos \dfrac{{c + d}}{2} . \sin \dfrac{{c - d}}{2}

⇒sin⁡47−sin⁡25=2cos⁡47+252.sin⁡47−252−−−−−(1) \Rightarrow \sin 47 - \sin 25 = 2\cos \dfrac{{47 + 25}}{2} . \sin \dfrac{{47 - 25}}{2} - - - - - (1)
⇒sin⁡61−sin⁡11=2cos⁡61+112.sin⁡61−112−−−−−(2)\Rightarrow \sin 61 - \sin 11 = 2\cos \dfrac{{61 + 11}}{2} . \sin \dfrac{{61 - 11}}{2} - - - - - (2)
Now, putting the value of (1) & (2) in the equation, we get,
(sin⁡47∘−sin⁡25∘)+(sin⁡61∘−sin⁡11∘)=2cos⁡47+252.sin⁡47−252+2cos⁡61+112.sin⁡61−112(\sin {47^ \circ } - \sin {25^ \circ }) + (\sin {61^ \circ } - \sin {11^ \circ }) = 2\cos \dfrac{{47 + 25}}{2} . \sin \dfrac{{47 - 25}}{2} + 2\cos \dfrac{{61 + 11}}{2} . \sin \dfrac{{61 - 11}}{2}
=2cos⁡722.sin⁡222+2cos⁡722.sin⁡502= 2\cos \dfrac{{72}}{2} . \sin \dfrac{{22}}{2} + 2\cos \dfrac{{72}}{2} . \sin \dfrac{{50}}{2}
=2cos⁡36.sin⁡11+2cos⁡36.sin⁡25= 2\cos 36 . \sin 11 + 2\cos 36 . \sin 25
Taking 2cos⁡362\cos 36 common, we get
=2cos⁡36(sin⁡11+sin⁡25)= 2\cos 36(\sin 11 + \sin 25)
=2cos⁡36(sin⁡25+sin⁡11)= 2\cos 36(\sin 25 + \sin 11)
Now, we use the formula sin⁡c+sin⁡d=2sin⁡c+d2.cos⁡c−d2\sin c + \sin d = 2\sin \dfrac{{c + d}}{2} . \cos \dfrac{{c - d}}{2}, we get
=2cos⁡36(2sin⁡25+112.cos⁡25−112)= 2\cos 36(2\sin \dfrac{{25 + 11}}{2} . \cos \dfrac{{25 - 11}}{2})
=4cos⁡36.sin⁡18.cos⁡7= 4\cos 36 . \sin 18 . \cos 7
Here, the value of cos⁡36=5+14\cos 36 = \dfrac{{\sqrt 5 + 1}}{4} and sin⁡18=5−14\sin 18 = \dfrac{{\sqrt 5 - 1}}{4}
Put the values in the equation we get,
=45+14.5−14.cos⁡7= 4\dfrac{{\sqrt 5 + 1}}{4} . \dfrac{{\sqrt 5 - 1}}{4} . \cos 7
We know that (a+b)(a−b)=a2−b2(a+b) (a-b)=a^2-b^2
So,
=4.(5)2−124×4.cos⁡7= 4 . \dfrac{{({\sqrt5})^2 - 1^2}}{{4 \times 4}} . \cos 7
=4.5−14×4.cos⁡7= 4 . \dfrac{{5 - 1}}{{4 \times 4}} . \cos 7
=cos⁡7∘= \cos {7^ \circ }
∴\therefore The value of sin⁡47∘−sin⁡25∘+sin⁡61∘−sin⁡11∘=cos⁡7∘\sin {47^ \circ } - \sin {25^ \circ } + \sin {61^ \circ } - \sin {11^ \circ } = \cos {7^ \circ }

So, the correct answer is “Option A”.

Additional Information:

  1. Sine and cosine are the functions enlightening the shape of a right angled triangle.
  2. Sine angle is the ratio of the side opposite to the vertex from which we are observing to the hypotenuse.
  3. Cosine angle is the ratio of the adjacent side of the vertex from where we are observing to the hypotenuse.
  4. Size of the triangle does not affect the values of sine angle and cosine angle for the given value of the angle.

Note: Students should remember the trigonometric sum and difference formulas i.e sin⁡c−sin⁡d=2cos⁡c+d2.sin⁡c−d2\sin c - \sin d = 2\cos \dfrac{{c + d}}{2} . \sin \dfrac{{c - d}}{2} and sin⁡c+sin⁡d=2sin⁡c+d2.cos⁡c−d2\sin c + \sin d = 2\sin \dfrac{{c + d}}{2} . \cos \dfrac{{c - d}}{2} and also some trigonometric standard angles cos⁡36=5+14\cos 36 = \dfrac{{\sqrt 5 + 1}}{4} and sin⁡18=5−14\sin 18 = \dfrac{{\sqrt 5 - 1}}{4} for solving these types of problems.