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Question: The value of \(\sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = \) A. \(\sin 36^\ci...

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

We can simplify the 1st with last term and 2nd term with 3rd term using the equation, sin⁡(A)−sin⁡(B)=2cos⁡(A+B2)sin⁡(A−B2)\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right). Then we can take the common factors and simplify them using the equationsin⁡(A)+sin⁡(B)=2sin⁡(A+B2)cos⁡(A−B2)\sin \left( A \right) + \sin \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right). Then we can find the values of the function at that angles and obtain the required answer.

Complete step by step answer:

We have the expression sin⁡47∘+sin⁡61∘−sin⁡11∘−sin⁡25∘\sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ
We can rearrange it as sin⁡47∘−sin⁡25∘+sin⁡61∘−sin⁡11∘\sin 47^\circ - \sin 25^\circ + \sin 61^\circ - \sin 11^\circ .
We can take the 1st two terms. sin⁡47∘−sin⁡25∘\sin 47^\circ - \sin 25^\circ
We know that, sin⁡(A)−sin⁡(B)=2cos⁡(A+B2)sin⁡(A−B2)\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) .
On substituting the values, we get,
⇒sin⁡47∘−sin⁡25∘=2cos⁡(47∘+25∘2)sin⁡(47∘−25∘2)\Rightarrow \sin 47^\circ - \sin 25^\circ = 2\cos \left( {\dfrac{{47^\circ + 25^\circ }}{2}} \right)\sin \left( {\dfrac{{47^\circ - 25^\circ }}{2}} \right).
On further calculation, we get,
⇒sin⁡47∘−sin⁡25∘=2cos⁡36∘sin⁡11∘\Rightarrow \sin 47^\circ - \sin 25^\circ = 2\cos 36^\circ \sin 11^\circ … (1)
We can take the last two terms. sin⁡61∘−sin⁡11∘\sin 61^\circ - \sin 11^\circ
We know that, sin⁡(A)−sin⁡(B)=2cos⁡(A+B2)sin⁡(A−B2)\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right) .
On substituting the values, we get,
⇒sin⁡61∘−sin⁡11∘=2cos⁡(61∘+11∘2)sin⁡(61∘−11∘2)\Rightarrow \sin 61^\circ - \sin 11^\circ = 2\cos \left( {\dfrac{{61^\circ + 11^\circ }}{2}} \right)\sin \left( {\dfrac{{61^\circ - 11^\circ }}{2}} \right).
On further calculation, we get,
⇒sin⁡61∘−sin⁡11∘=2cos⁡36∘sin⁡25∘\Rightarrow \sin 61^\circ - \sin 11^\circ = 2\cos 36^\circ \sin 25^\circ … (2)
Substituting (1) and (2) in expression, we get,
⇒l=2cos⁡36∘sin⁡11∘+2cos⁡36∘sin⁡25∘\Rightarrow l = 2\cos 36^\circ \sin 11^\circ + 2\cos 36^\circ \sin 25^\circ
Let l=sin⁡47∘+sin⁡61∘−sin⁡11∘−sin⁡25∘l = \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ ,
⇒l=2cos⁡36∘sin⁡11∘+2cos⁡36∘sin⁡25∘\Rightarrow l = 2\cos 36^\circ \sin 11^\circ + 2\cos 36^\circ \sin 25^\circ
We can take 2cos⁡36∘2\cos 36^\circ common,
⇒I=2cos⁡36∘(sin⁡11∘+sin⁡25∘)\Rightarrow I = 2\cos 36^\circ \left( {\sin 11^\circ + \sin 25^\circ } \right)
We know thatsin⁡(A)+sin⁡(B)=2sin⁡(A+B2)cos⁡(A−B2)\sin \left( A \right) + \sin \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right).
⇒sin⁡11∘+sin⁡25∘=2sin⁡(11∘+25∘2)cos⁡(11∘−25∘2)\Rightarrow \sin 11^\circ + \sin 25^\circ = 2\sin \left( {\dfrac{{11^\circ + 25^\circ }}{2}} \right)\cos \left( {\dfrac{{11^\circ - 25^\circ }}{2}} \right)
On further calculation, we get,
⇒sin⁡11∘+sin⁡25∘=2sin⁡18∘cos⁡(−7∘)\Rightarrow \sin 11^\circ + \sin 25^\circ = 2\sin 18^\circ \cos \left( { - 7^\circ } \right)
We know that cos⁡(−x)=cos⁡(x)\cos \left( { - x} \right) = \cos \left( x \right)
⇒sin⁡11∘+sin⁡25∘=2sin⁡18∘cos⁡7∘\Rightarrow \sin 11^\circ + \sin 25^\circ = 2\sin 18^\circ \cos 7^\circ
Substituting this in the expression, we get,
⇒I=2cos⁡36∘(2sin⁡18∘cos⁡7∘)\Rightarrow I = 2\cos 36^\circ \left( {2\sin 18^\circ \cos 7^\circ } \right)
Now we can apply the value of 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},
⇒I=4×5+14×5−14×cos⁡7∘\Rightarrow I = 4 \times \dfrac{{\sqrt 5 + 1}}{4} \times \dfrac{{\sqrt 5 - 1}}{4} \times \cos 7^\circ
We apply the identity, (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2},
⇒I=5−14×cos⁡7∘\Rightarrow I = \dfrac{{5 - 1}}{4} \times \cos 7^\circ
⇒I=cos⁡7∘\Rightarrow I = \cos 7^\circ
Therefore the value of the expression is cos⁡7∘\cos 7^\circ
So the correct answer is option D.

Note: We must be familiar to the following trigonometric identities used in this problem.
cos⁡(A)+cos⁡(B)=2cos⁡(A+B2)cos⁡(A−B2)\cos \left( A \right) + \cos \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)
cos⁡(A)−cos⁡(B)=−2sin⁡(A+B2)sin⁡(A−B2)\cos \left( A \right) - \cos \left( B \right) = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
sin⁡(A)+sin⁡(B)=2sin⁡(A+B2)cos⁡(A−B2)\sin \left( A \right) + \sin \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)
sin⁡(A)−sin⁡(B)=2cos⁡(A+B2)sin⁡(A−B2)\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
sin⁡(−x)=−sin⁡(x)\sin \left( { - x} \right) = - \sin \left( x \right)
cos⁡(−x)=cos⁡(x)\cos \left( { - x} \right) = \cos \left( x \right)
We must know the values of trigonometric functions at common angles. Adding π\pi or multiples of π\pi with the angle retains the ratio and adding π2\dfrac{\pi }{2}or odd multiples of π2\dfrac{\pi }{2}will change the ratio. While converting the angles we must take care of the sign of the ratio in its respective quadrant. In the 1st quadrant all the trigonometric ratios are positive. In the 2nd quadrant only sine and sec are positive. In the third quadrant, only tan and cot are positive and in the fourth quadrant, only cos and sec are positive. The angle measured in the counter clockwise direction is taken as positive and angle measured in the clockwise direction is taken as negative.