Prove that:

Solution:

(i) Consider LHS :

$\cos 55^{\circ}+\cos 65^{\circ}+\cos 175^{\circ}$

$=2 \cos \left(\frac{55^{\circ}+65^{\circ}}{2}\right) \cos \left(\frac{55^{\circ}-65^{\circ}}{2}\right)+\cos 175^{\circ} \quad\left\{\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \cos 60^{\circ} \cos \left(-5^{\circ}\right)+\cos 175^{\circ}$

$=2 \times \frac{1}{2} \cos 5^{\circ}+\cos 175^{\circ}$

$=\cos 5^{\circ}+\cos 175^{\circ}$

$=2 \cos \left(\frac{5^{\circ}+175^{\circ}}{2}\right) \cos \left(\frac{5^{\circ}-175^{\circ}}{2}\right)$

$=2 \cos 90^{\circ} \cos 85^{\circ}$

$=0$

Hence, LHS = RHS.

(ii) Consider LHS :

$\sin 50^{\circ}-\sin 70^{\circ}+\sin 10^{\circ}$

$=2 \sin \left(\frac{50^{\circ}-70^{\circ}}{2}\right) \cos \left(\frac{50^{\circ}+70^{\circ}}{2}\right)+\sin 10^{\circ} \quad\left\{\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) c \cos \left(\frac{A+B}{2}\right)\right\}$

$=2 \sin \left(-10^{\circ}\right) \cos 60^{\circ}+\sin 10^{\circ}$

$=2 \times \frac{1}{2} \sin \left(-10^{\circ}\right)+\sin 10^{\circ}$

$=-\sin 10^{\circ}+\sin 10^{\circ}$

$=0$

Hence, LHS = RHS.

(iii) Consider LHS :

$\cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}$

$=2 \cos \left(\frac{80^{\circ}+40^{\circ}}{2}\right) \cos \left(\frac{80^{\circ}-40^{\circ}}{2}\right)-\cos 20^{\circ} \quad\left\{\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \cos 60^{\circ} \cos 20^{\circ}-\cos 20^{\circ}$

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

$=\cos 20^{\circ}-\cos 20^{\circ}$

$=0$

Hence, LHS = RHS.

(iv) Consider LHS :

$\cos 20^{\circ}+\cos 100^{\circ}+\cos 140^{\circ}$

$=2 \cos \left(\frac{20^{\circ}+100^{\circ}}{2}\right) \cos \left(\frac{20^{\circ}-100^{\circ}}{2}\right)+\cos 140^{\circ} \quad\left\{\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \cos 60^{\circ} \cos \left(-40^{\circ}\right)+\cos 140^{\circ}$

$=2 \times \frac{1}{2} \cos 40^{\circ}+\cos 140^{\circ}$

$=\cos 40^{\circ}+\cos 140^{\circ}$

$=2 \cos \left(\frac{40^{\circ}+140^{\circ}}{2}\right) \cos \left(\frac{40^{\circ}-140^{\circ}}{2}\right)$

$=2 \cos 90^{\circ} \cos 50^{\circ}$

$=0$

Hence, LHS = RHS.

(v) Consider LHS :

$\mathrm{LHS}=\sin \left(\frac{5 \pi}{18}\right)-\cos \frac{4 \pi}{9}$

$=\sin \left(\frac{5 \pi}{18}\right)-\cos \left(\frac{\pi}{2}-\frac{\pi}{18}\right)$

$=\sin \left(\frac{5 \pi}{18}\right)-\sin \left(\frac{\pi}{18}\right)$

$=2 \sin \left(\frac{\frac{5 \pi}{18}-\frac{\pi}{18}}{2}\right) \cos \left(\frac{\frac{5 \pi}{18}+\frac{\pi}{18}}{2}\right) \quad\left[\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right]$

$=2 \sin \left(\frac{\pi}{9}\right) \cos \frac{\pi}{6}$

$=2 \sin \left(\frac{\pi}{9}\right) \cos \frac{\pi}{6}$

$=2 \times \frac{\sqrt{3}}{2} \sin \left(\frac{\pi}{9}\right)$

$=\sqrt{3} \sin \left(\frac{\pi}{9}\right)=\mathrm{RHS}$

Hence, LHS = RHS.

(vi) Consider LHS :

LHS $=\cos \frac{\pi}{12}-\sin \frac{\pi}{12}$

$=\cos \left(\frac{\pi}{2}-\frac{5 \pi}{12}\right)-\sin \frac{\pi}{12}$

$=\sin \left(\frac{5 \pi}{12}\right)-\sin \frac{\pi}{12}$$=\sin \left(\frac{5 \pi}{12}\right)-\sin \frac{\pi}{12}$

$=2 \sin \left(\frac{\frac{5 \pi}{12}-\frac{\pi}{12}}{2}\right) \cos \left(\frac{\frac{5 \pi}{12}+\frac{\pi}{12}}{2}\right) \quad\left\{\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right\}$

$=2 \sin \left(\frac{\pi}{6}\right) \cos \left(\frac{\pi}{4}\right)$

$=2 \times \frac{1}{2} \times \frac{1}{\sqrt{2}}$

$=\frac{1}{\sqrt{2}}$

Hence, LHS = RHS.

(vii) Consider LHS :

$s \operatorname{in} 80^{\circ}-\cos 70^{\circ}$

$=\sin 80^{\circ}-\cos \left(90^{\circ}-20^{\circ}\right)$

$=\sin 80^{\circ}-\sin 20^{\circ}$

$=2 \sin \left(\frac{80^{\circ}-20^{\circ}}{2}\right) \cos \left(\frac{80^{\circ}+20^{\circ}}{2}\right) \quad\left\{\because \sin A-\sin B=2 \sin \left(\frac{A-B}{2}\right) \cos \left(\frac{A+B}{2}\right)\right\}$

$=2 \sin 30^{\circ} \cos 50^{\circ}$

$=2 \times \frac{1}{2} \cos 50^{\circ}$

$=\cos 50^{\circ}$

$=\operatorname{RH} S$

Hence, LHS = RHS.

(viii) Consider LHS :

$\sin 51^{\circ}+\cos 81^{\circ}$

$=\sin 51^{\circ}+\cos \left(90^{\circ}-9^{\circ}\right)$

$=\sin 51^{\circ}+\sin 9^{\circ}$

$=2 \sin \left(\frac{51^{\circ}+9^{\circ}}{2}\right) \cos \left(\frac{51^{\circ}-9^{\circ}}{2}\right) \quad\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \sin 30^{\circ} \cos 21^{\circ}$

$=2 \times \frac{1}{2} \cos \left(21^{\circ}\right)$

$=\cos \left(21^{\circ}\right)$

$=\mathrm{RHS}$

Hence, LHS = RHS.