Question

Find the value of $\sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ$

Answer 1

Here, we will first rewrite the given expression and use the sum of sines in terms of products. We will then simplify it further and use the difference of sines in terms of products. Then we will use the algebraic identity and simplify the given trigonometric expression to get the required answer.

Formula Used:
We will use the following formula:

1. The sum of sines in terms of products is given by the formula $\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$
2. The Difference of sines in terms of products is given by the formula $\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
3. The difference between the square of the numbers is given by the algebraic identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$

Complete step by step solution:
We are given with a Trigonometric function $\sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ$
Rewriting the giving expression, we get
$\Rightarrow \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = \sin 47^\circ + \sin 61^\circ - \left( {\sin 11^\circ + \sin 25^\circ } \right)$
Now, by using the sines in terms of products formula $\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$, we get
$\Rightarrow \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 2\sin \left( {\dfrac{{47^\circ + 61^\circ }}{2}} \right)\cos \left( {\dfrac{{47^\circ - 61^\circ }}{2}} \right) - \left[ {2\sin \left( {\dfrac{{11^\circ + 25^\circ }}{2}} \right)\cos \left( {\dfrac{{11^\circ - 25^\circ }}{2}} \right)} \right]$ Now, by simplifying the equation, we get
$\Rightarrow \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 2\sin \left( {\dfrac{{108^\circ }}{2}} \right)\cos \left( {\dfrac{{ - 14^\circ }}{2}} \right) - \left[ {2\sin \left( {\dfrac{{36^\circ }}{2}} \right)\cos \left( {\dfrac{{ - 14^\circ }}{2}} \right)} \right]$
$\Rightarrow \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 2\sin \left( {54^\circ } \right)\cos \left( { - 7^\circ } \right) - \left[ {2\sin \left( {18^\circ } \right)\cos \left( { - 7^\circ } \right)} \right]$
We know that $\cos \left( { - \theta } \right) = \cos \theta$ . So, we get
$\Rightarrow \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 2\sin \left( {54^\circ } \right)\cos \left( {7^\circ } \right) - \left[ {2\sin \left( {18^\circ } \right)\cos \left( {7^\circ } \right)} \right]$
Now, by taking out the common factor, we get
$\Rightarrow \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 2\cos 7^\circ \left[ {\sin \left( {54^\circ } \right) - \sin \left( {18^\circ } \right)} \right]$
The Difference of sines in terms of products is given by the formula $\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
Now, by using the sines in terms of products formula, we get
$\Rightarrow \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 2\cos 7^\circ \left[ {2\cos \left( {\dfrac{{54^\circ + 18^\circ }}{2}} \right)\sin \left( {\dfrac{{54^\circ - 18^\circ }}{2}} \right)} \right]$
Now, by simplifying the equation, we get
$\Rightarrow \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 2\cos 7^\circ \left[ {2\cos \left( {\dfrac{{72^\circ }}{2}} \right)\sin \left( {\dfrac{{36^\circ }}{2}} \right)} \right]$
$\Rightarrow \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 2\cos 7^\circ \left[ {2\cos \left( {36^\circ } \right)\sin \left( {18^\circ } \right)} \right]$
We know that $\sin 18^\circ = \dfrac{{\sqrt 5 - 1}}{4}$ and $\cos 36^\circ = \dfrac{{\sqrt 5 + 1}}{4}$
Now, by substituting the known values, we get
$\Rightarrow \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 4\cos 7^\circ \times \dfrac{{\sqrt 5 - 1}}{4} \times \dfrac{{\sqrt 5 + 1}}{4}$
The difference between the square of the numbers is given by the algebraic identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
Now, by using the algebraic identity, we get
$\Rightarrow \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 4\cos 7^\circ \times \dfrac{{{{\left( {\sqrt 5 } \right)}^2} - {1^2}}}{{4 \times 4}}$
Applying the exponent on the terms, we get
$\Rightarrow \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 4\cos 7^\circ \times \dfrac{{5 - 1}}{{4 \times 4}}$
$\Rightarrow \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = 4\cos 7^\circ \times \dfrac{4}{{4 \times 4}}$
Now, by simplifying the equation, we get
$\Rightarrow \sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = \cos 7^\circ$

Therefore, the value of $\sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ$ is $\cos 7^\circ$.

Note:
We know that Trigonometric Equation is defined as an equation involving the trigonometric ratios. Trigonometric identity is an equation which is always true for all the variables. We should note in particular that sine and tangent are odd functions since both the functions are symmetric about the origin. Cosine is an even function because the function is symmetric about y axis. So, we take the arguments in the negative sign for odd functions and positive signs for even functions. Trigonometric Ratios of a Particular angle are the ratios of the sides of a right angled triangle with respect to any of its acute angle.