Question

The value of $\sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = $
A. $\sin 36^\circ $
B. $\cos 36^\circ $
C. $\sin 7^\circ $
D. $\cos 7^\circ $

Answer 1

Hint: We can simplify the 1st with last term and 2nd term with 3rd term using the equation, \[\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)\]. Then we can take the common factors and simplify them using the equation\[\sin \left( A \right) + \sin \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\]. 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^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ $
We can rearrange it as $\sin 47^\circ - \sin 25^\circ + \sin 61^\circ - \sin 11^\circ $.
We can take the 1st two terms. $\sin 47^\circ - \sin 25^\circ $
We know that, \[\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)\] .
On substituting the values, we get,
\[ \Rightarrow \sin 47^\circ - \sin 25^\circ = 2\cos \left( {\dfrac{{47^\circ + 25^\circ }}{2}} \right)\sin \left( {\dfrac{{47^\circ - 25^\circ }}{2}} \right)\].
On further calculation, we get,
\[ \Rightarrow \sin 47^\circ - \sin 25^\circ = 2\cos 36^\circ \sin 11^\circ \] … (1)
We can take the last two terms. $\sin 61^\circ - \sin 11^\circ $
We know that, \[\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)\] .
On substituting the values, we get,
\[ \Rightarrow \sin 61^\circ - \sin 11^\circ = 2\cos \left( {\dfrac{{61^\circ + 11^\circ }}{2}} \right)\sin \left( {\dfrac{{61^\circ - 11^\circ }}{2}} \right)\].
On further calculation, we get,
\[ \Rightarrow \sin 61^\circ - \sin 11^\circ = 2\cos 36^\circ \sin 25^\circ \] … (2)
Substituting (1) and (2) in expression, we get,
$ \Rightarrow l = 2\cos 36^\circ \sin 11^\circ + 2\cos 36^\circ \sin 25^\circ $
Let $l = \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ $,
$ \Rightarrow l = 2\cos 36^\circ \sin 11^\circ + 2\cos 36^\circ \sin 25^\circ $
We can take $2\cos 36^\circ $ common,
\[ \Rightarrow I = 2\cos 36^\circ \left( {\sin 11^\circ + \sin 25^\circ } \right)\]
We know that\[\sin \left( A \right) + \sin \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\].
\[ \Rightarrow \sin 11^\circ + \sin 25^\circ = 2\sin \left( {\dfrac{{11^\circ + 25^\circ }}{2}} \right)\cos \left( {\dfrac{{11^\circ - 25^\circ }}{2}} \right)\]
On further calculation, we get,
\[ \Rightarrow \sin 11^\circ + \sin 25^\circ = 2\sin 18^\circ \cos \left( { - 7^\circ } \right)\]
We know that \[\cos \left( { - x} \right) = \cos \left( x \right)\]
\[ \Rightarrow \sin 11^\circ + \sin 25^\circ = 2\sin 18^\circ \cos 7^\circ \]
Substituting this in the expression, we get,
\[ \Rightarrow I = 2\cos 36^\circ \left( {2\sin 18^\circ \cos 7^\circ } \right)\]
Now we can apply the value of $\cos 36^\circ = \dfrac{{\sqrt 5 + 1}}{4}$ and $\sin 18^\circ = \dfrac{{\sqrt 5 - 1}}{4}$,
\[ \Rightarrow I = 4 \times \dfrac{{\sqrt 5 + 1}}{4} \times \dfrac{{\sqrt 5 - 1}}{4} \times \cos 7^\circ \]
We apply the identity, $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$,
\[ \Rightarrow I = \dfrac{{5 - 1}}{4} \times \cos 7^\circ \]
\[ \Rightarrow I = \cos 7^\circ \]
Therefore the value of the expression is \[\cos 7^\circ \]
So the correct answer is option D.

Note: We must be familiar to the following trigonometric identities used in this problem.
\[\cos \left( A \right) + \cos \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\]
\[\cos \left( A \right) - \cos \left( B \right) = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)\]
\[\sin \left( A \right) + \sin \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\]
\[\sin \left( A \right) - \sin \left( B \right) = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)\]
\[\sin \left( { - x} \right) = - \sin \left( x \right)\]
\[\cos \left( { - x} \right) = \cos \left( x \right)\]
We must know the values of trigonometric functions at common angles. Adding $\pi $or multiples of $\pi $with the angle retains the ratio and adding $\dfrac{\pi }{2}$or odd multiples of $\dfrac{\pi }{2}$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.