47^ + 61^ - 11^ - 25^ is equal to

MCQ

$\sin47^\circ+\sin61^\circ-\sin11^\circ-\sin25^\circ$ is equal to

A
$\sin36^\circ$
B
$\cos36^\circ$
C
$\sin7^\circ$
✓
$\cos7^\circ$

✓

Answer

Correct option: D.

$\cos7^\circ$

$\sin47^\circ+\sin61^\circ-\sin11^\circ-\sin25^\circ$
$=\ \sin47^\circ-\sin25^\circ+\sin61^\circ-\sin11^\circ$
$=\ 2\sin\Big(\frac{47^\circ-25^\circ}{2}\Big)\cos\Big(\frac{47^\circ+25^\circ}{2}\Big)\\ \ \ \ +2\sin\Big(\frac{61^\circ-11^\circ}{2}\Big)\cos\Big(\frac{61^\circ+11^\circ}{2}\Big)$
$=\ 2\sin11^\circ\cos36^\circ+2\sin25^\circ\cos36^\circ$
$=\ 2\cos36^\circ(\sin11^\circ+\sin25^\circ)$
$=\ 2\cos36^\circ\Big\{2\sin\Big(\frac{11^\circ+25^\circ}{2}\Big)\cos\Big(\frac{11^\circ-25^\circ}{2}\Big)\Big\}$
$=\ 4\cos36^\circ\sin18^\circ\cos7^\circ$
$=\ 4\times\Big(\frac{\sqrt5-1}{4}\Big)\Big(\frac{\sqrt5+1}{4}\Big)\cos7^\circ$ $\Big[\cos36^\circ=\frac{\sqrt5+1}{4}\text{ and }\sin18^\circ=\frac{\sqrt5-1}{4}\Big]$
$=\ \frac{5-1}{4}\cos7^\circ$
$=\ \cos7^\circ$

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