Prove that:

Solution:

(i) Consider LHS :

$\sin 65^{\circ}+\cos 65^{\circ}$

$=\sin 65^{\circ}+\cos \left(90^{\circ}-25^{\circ}\right)$

$=\sin 65^{\circ}+\sin 25^{\circ}$

$=2 \sin \left(\frac{65^{\circ}+25^{\circ}}{2}\right) \cos \left(\frac{65^{\circ}-25^{\circ}}{2}\right) \quad\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \sin 45^{\circ} \cos 20^{\circ}$

$=2 \times \frac{1}{\sqrt{2}} \cos 20^{\circ}$

$=\sqrt{2} \cos 20^{\circ}$

$=\mathrm{RHS}$

Hence, LHS $=$ RHS.

(i) Consider LHS :

$\sin 47^{\circ}+\cos 77^{\circ}$$=\sin 47^{\circ}+\cos \left(90^{\circ}-13^{\circ}\right)$

$=\sin 47^{\circ}+\sin 13^{\circ}$

$=2 \sin \left(\frac{47^{\circ}+13^{\circ}}{2}\right) \cos \left(\frac{47^{\circ}-13^{\circ}}{2}\right) \quad\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \sin 30^{\circ} \cos 17^{\circ}$

$=2 \times \frac{1}{2} \cos 17^{\circ}$

$=\cos 17^{\circ}$

$=\mathrm{RHS}$

Hence, LHS $=$ RHS.