# sin 47° + sin 61° − sin 11° − sin 25° is equal to

Question:

sin 47° + sin 61° − sin 11° − sin 25° is equal to

(a) sin 36°

(b) cos 36°

(c) sin 7°

(d) cos 7°

Solution:

(d) cos 7°

$\sin 47^{\circ}+\sin 61^{\circ}-\sin 11^{\circ}-\sin 25^{\circ}$

$=\sin 47^{\circ}-\sin 25^{\circ}+\sin 61^{\circ}-\sin 11^{\circ}$

$=2 \sin \left(\frac{47^{\circ}-25^{\circ}}{2}\right) \cos \left(\frac{47^{\circ}+25^{\circ}}{2}\right)+2 \sin \left(\frac{61^{\circ}-11^{\circ}}{2}\right) \cos \left(\frac{61^{\circ}+11^{\circ}}{2}\right)$

$=2 \sin 11^{\circ} \cos 36^{\circ}+2 \sin 25^{\circ} \cos 36^{\circ}$

$=2 \cos 36^{\circ}\left(\sin 11^{\circ}+\sin 25^{\circ}\right)$

$=2 \cos 36^{\circ}\left\{2 \sin \left(\frac{11^{\circ}+25^{\circ}}{2}\right) \cos \left(\frac{11^{\circ}-25^{\circ}}{2}\right)\right\}$

$=4 \cos 36^{\circ} \sin 18^{\circ} \cos 7^{\circ}$

$=4 \times\left(\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}\right) \cos 7^{\circ} \quad\left[\cos 36^{\circ}=\frac{\sqrt{5}+1}{4}\right.$ and $\left.\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}\right]$

$=\frac{5-1}{4} \cos 7^{\circ}$

$=\cos 7^{\circ}$