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

Last updated date: 20th Jun 2024
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Hint:According to the question we have to find the value of the following equation $\sin {47^ \circ } + \sin {61^ \circ } - \sin {11^ \circ } - \sin {25^ \circ }$. To determine the value of the equation we use the formula $\sin c - \sin d = 2\cos \dfrac{{c + d}}{2} . \sin \dfrac{{c - d}}{2}$. Then put values in place of c and d we get the equation $(\sin {47^ \circ } - \sin {25^ \circ }) + (\sin {61^ \circ } - \sin {11^ \circ }) = 2\cos \dfrac{{47 + 25}}{2} . \sin \dfrac{{47 - 25}}{2} + 2\cos \dfrac{{61 + 11}}{2} . \sin \dfrac{{61 - 11}}{2}$.By solving this we get the value of the equation.

In the question we will find the value of $\sin {47^ \circ } + \sin {61^ \circ } - \sin {11^ \circ } - \sin {25^ \circ }$
First we arrange the equation in the following form $\sin {47^ \circ } - \sin {25^ \circ } + \sin {61^ \circ } - \sin {11^ \circ }$
For finding the value we use the formula
$\to \sin c - \sin d = 2\cos \dfrac{{c + d}}{2} . \sin \dfrac{{c - d}}{2}$

$\Rightarrow \sin 47 - \sin 25 = 2\cos \dfrac{{47 + 25}}{2} . \sin \dfrac{{47 - 25}}{2} - - - - - (1)$
$\Rightarrow \sin 61 - \sin 11 = 2\cos \dfrac{{61 + 11}}{2} . \sin \dfrac{{61 - 11}}{2} - - - - - (2)$
Now, putting the value of (1) & (2) in the equation, we get,
$(\sin {47^ \circ } - \sin {25^ \circ }) + (\sin {61^ \circ } - \sin {11^ \circ }) = 2\cos \dfrac{{47 + 25}}{2} . \sin \dfrac{{47 - 25}}{2} + 2\cos \dfrac{{61 + 11}}{2} . \sin \dfrac{{61 - 11}}{2}$
$= 2\cos \dfrac{{72}}{2} . \sin \dfrac{{22}}{2} + 2\cos \dfrac{{72}}{2} . \sin \dfrac{{50}}{2}$
$= 2\cos 36 . \sin 11 + 2\cos 36 . \sin 25$
Taking $2\cos 36$ common, we get
$= 2\cos 36(\sin 11 + \sin 25)$
$= 2\cos 36(\sin 25 + \sin 11)$
Now, we use the formula $\sin c + \sin d = 2\sin \dfrac{{c + d}}{2} . \cos \dfrac{{c - d}}{2}$, we get
$= 2\cos 36(2\sin \dfrac{{25 + 11}}{2} . \cos \dfrac{{25 - 11}}{2})$
$= 4\cos 36 . \sin 18 . \cos 7$
Here, the value of $\cos 36 = \dfrac{{\sqrt 5 + 1}}{4}$ and $\sin 18 = \dfrac{{\sqrt 5 - 1}}{4}$
Put the values in the equation we get,
$= 4\dfrac{{\sqrt 5 + 1}}{4} . \dfrac{{\sqrt 5 - 1}}{4} . \cos 7$
We know that $(a+b) (a-b)=a^2-b^2$
So,
$= 4 . \dfrac{{({\sqrt5})^2 - 1^2}}{{4 \times 4}} . \cos 7$
$= 4 . \dfrac{{5 - 1}}{{4 \times 4}} . \cos 7$
$= \cos {7^ \circ }$
$\therefore$ The value of $\sin {47^ \circ } - \sin {25^ \circ } + \sin {61^ \circ } - \sin {11^ \circ } = \cos {7^ \circ }$

So, the correct answer is “Option A”.

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