Prove that:

Solution:

(i) Consider LHS :

$s$ in $38^{\circ}+\sin 22^{\circ}$

$=2 \sin \left(\frac{38^{\circ}+22^{\circ}}{2}\right) \cos \left(\frac{38^{\circ}-22^{\circ}}{2}\right)$   $\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \sin 30^{\circ} \cos 8^{\circ}$

$=2 \times \frac{1}{2} \cos \left(90^{\circ}-8^{\circ}\right)$

$=\sin 82^{\circ}$

= RHS

Hence, LHS = RHS.

(ii) Consider LHS :

$\cos 100^{\circ}+\cos 20^{\circ}$

$=2 \cos \left(\frac{100^{\circ}+20^{\circ}}{2}\right) \cos \left(\frac{100^{\circ}-20^{\circ}}{2}\right) \quad\left\{\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \cos 60^{\circ} \cos 40^{\circ}$

$=2 \times \frac{1}{2} \cos 40^{\circ}$

$=\cos 40^{\circ}$

Hence, LHS = RHS.

(iii) Consider LHS :

$\sin 50^{\circ}+\sin 10^{\circ}$

$=2 \sin \left(\frac{50^{\circ}+10^{\circ}}{2}\right) \cos \left(\frac{50^{\circ}-10^{\circ}}{2}\right) \quad\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \sin 30^{\circ} \cos 20^{\circ}$

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

$=\cos 20^{\circ}$

Hence, LHS = RHS.

(iv) Consider LHS :

$s$ in $23^{\circ}+\sin 37^{\circ}$

$=2 \sin \left(\frac{23^{\circ}+37^{\circ}}{2}\right) \cos \left(\frac{23^{\circ}-37^{\circ}}{2}\right) \quad\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \sin 30^{\circ} \cos \left(-7^{\circ}\right)$

$=2 \sin 30^{\circ} \cos 7^{\circ}$

$=2 \times \frac{1}{2} \cos 7^{\circ}$

$=\cos 7^{\circ}$

Hence, LHS = RHS.

(v) Consider LHS :

$s$ in $105^{\circ}+\cos 105^{\circ}$

$=\sin 105^{\circ}+\cos \left(90^{\circ}+15^{\circ}\right)$

$=\sin 105^{\circ}-\sin 15^{\circ}$

$=2 \sin \left(\frac{105^{\circ}-15^{\circ}}{2}\right) \cos \left(\frac{105^{\circ}+15^{\circ}}{2}\right) \quad\left\{\because \sin A+\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right\}$

$=2 \sin 45^{\circ} \cos 60^{\circ}$

$=2 \sin \left(90^{\circ}-45^{\circ}\right) \cos 60^{\circ}$

$=2 \times \frac{1}{2} \cos \left(45^{\circ}\right)$

$=\cos 45^{\circ}$

Hence, LHS = RHS.

(vi) Consider LHS :

$\sin 40^{\circ}+\sin 20^{\circ}$

$=2 \sin \left(\frac{40^{\circ}+20^{\circ}}{2}\right) \cos \left(\frac{40^{\circ}-20^{\circ}}{2}\right) \quad\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \sin 30^{\circ} \cos 10^{\circ}$

$=2 \times \frac{1}{2} \cos 10^{\circ}$

$=\cos \left(10^{\circ}\right)$

Hence, LHS = RHS.