# Prove that:

Question:

Prove that:

(i) cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A

(ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A

(iii) $\sin A+\sin 2 A+\sin 4 A+\sin 5 A=4 \cos \frac{A}{2} \cos \frac{3 A}{2} \sin 3 A$

(iv) $\sin 3 A+\sin 2 A-\sin A=4 \sin A \cos \frac{A}{2} \cos \frac{3 A}{2}$

(v) $\cos 20^{\circ} \cos 100^{\circ}+\cos 100^{\circ} \cos 140^{\circ}-140^{\circ} \cos 200^{\circ}=-\frac{3}{4}$

(vi) $\sin \frac{x}{2} \sin \frac{7 x}{2}+\sin \frac{3 x}{2} \sin \frac{11 x}{2}=\sin 2 x \sin 5 x$.

(vii) $\cos x \cos \frac{x}{2}-\cos 3 x \cos \frac{9 x}{2}=\sin 7 x \sin 8 x$

Solution:

(i) Consider LHS :

$\cos 3 A+c \cos 5 A+\cos 7 A+\cos 15 A$

$=2 \cos \left(\frac{3 A+5 A}{2}\right) \cos \left(\frac{3 A-5 A}{2}\right)+2 \cos \left(\frac{7 A+15 A}{2}\right) \cos \left(\frac{7 A-15 A}{2}\right)$

$\left\{\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \cos 4 A \cos (-A)+2 \cos 11 A \cos (-4 A)$

$=2 \cos 4 A \cos A+2 \cos 11 A \cos 4 A$

$=2 \cos 4 A\{\cos A+\cos 11 A\}$

$=2 \cos 4 A \times\left\{2 \cos \left(\frac{A+11 A}{2}\right) \cos \left(\frac{A-11 A}{2}\right)\right\}$

$=4 \cos 4 A \cos 6 A \cos (-5 A)$

$=4 \cos 4 A \cos 5 A \cos 6 A$

= RHS

Hence, LHS = RHS

(ii) Consider LHS :

$\cos A+\cos 3 A+\cos 5 A+\cos 7 A$

$=2 \cos \left(\frac{A+3 A}{2}\right) \cos \left(\frac{A-3 A}{2}\right)+2 \cos \left(\frac{5 A+7 A}{2}\right) \cos \left(\frac{5 A-7 A}{2}\right)$

$\left\{\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

$=2 \cos 2 A \cos (-A)+2 \cos 6 A \cos (-A)$

$=2 \cos 2 A \cos A+2 \cos 6 A \cos A$

$=2 \cos A(\cos 2 A+\cos 6 A)$

$=2 \cos A \times 2 \cos \left(\frac{2 A+6 A}{2}\right) \cos \left(\frac{2 A-6 A}{2}\right)$

$=4 \cos A \cos 4 A \cos (-2 A)$

$=4 \cos A \cos 2 A \cos 4 A$

= RHS

Hence, LHS = RHS

(iii) Consider LHS :

$\sin A+\sin 2 A+\sin 4 A+\sin 5 A$

$=2 \sin \left(\frac{A+2 A}{2}\right) \cos \left(\frac{A-2 A}{2}\right)+2 \sin \left(\frac{4 A+5 A}{2}\right) \cos \left(\frac{4 A-5 A}{2}\right)$

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

$=2 \sin \left(\frac{3}{2} A\right) \cos \left(-\frac{A}{2}\right)+2 \sin \left(\frac{9}{2} A\right) \cos \left(-\frac{A}{2}\right)$

$=2 \sin \left(\frac{3}{2} A\right) \cos \left(\frac{A}{2}\right)+2 \sin \left(\frac{9}{2} A\right) \cos \left(\frac{A}{2}\right)$

$=2 \cos \left(\frac{A}{2}\right)\left\{\sin \frac{3}{2} A+\sin \frac{9}{2} A\right\}$

$=2 \cos \left(\frac{A}{2}\right) \times 2 \sin \left(\frac{\frac{3}{2} A+\frac{9}{2} A}{2}\right) \cos \left(\frac{\frac{3}{2} A-\frac{9}{2} A}{2}\right)$

$=4 \cos \left(\frac{A}{2}\right) \cos 3 A \cos \left(-\frac{3}{2} A\right)$

$=4 \cos \frac{A}{2} \cos \left(\frac{3}{2} A\right) \cos 3 A$

= RHS

Hence, LHS = RHS

(iv) Consider LHS :

$=\sin 3 A+\sin 2 A-\sin A$

$=2 \sin \left(\frac{3 A+2 A}{2}\right) \cos \left(\frac{3 A-2 A}{2}\right)-\sin A \quad\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$

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

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

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

$=2 \cos \left(\frac{A}{2}\right) \times 2 \sin \left(\frac{\frac{5}{2} A-\frac{A}{2}}{2}\right) \cos \left(\frac{\frac{5}{2} A+\frac{A}{2}}{2}\right)$

$=4 \cos \left(\frac{A}{2}\right) \sin A \cos \left(\frac{3}{2} A\right)$

$=4 \sin A \cos \left(\frac{A}{2}\right) \cos \left(\frac{3}{2} A\right)$

= RHS

Hence, LHS = RHS

(v) Consider LHS :

$\cos 20^{\circ} \cos 100^{\circ}+\cos 100^{\circ} \cos 140^{\circ}-\cos 140^{\circ} \cos 200^{\circ}$

$=\frac{1}{2}\left(2 \cos 20^{\circ} \cos 100^{\circ}+2 \cos 100^{\circ} \cos 140^{\circ}-2 \cos 140^{\circ} \cos 200^{\circ}\right)$

$=\frac{1}{2}\left[\cos \left(100^{\circ}+20^{\circ}\right) \cos \left(100^{\circ}-20^{\circ}\right)+\cos \left(140^{\circ}+100^{\circ}\right) \cos \left(140^{\circ}-100^{\circ}\right)-\cos \left(200^{\circ}+140^{\circ}\right) \cos \left(200^{\circ}-140^{\circ}\right)\right]$

$=\frac{1}{0}\left[\cos 120^{\circ}+\cos 80^{\circ}+\cos 240^{\circ}+\cos 40^{\circ}-\cos 340^{\circ}-\cos 60^{\circ}\right]$

$=\frac{1}{2}\left[\cos 120^{\circ}+\cos 240^{\circ}-\cos 60^{\circ}+\cos 80^{\circ}+\cos 40^{\circ}-\cos 340^{\circ}\right]$

$=\frac{1}{2}\left[\left(-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\right)+\cos 80^{\circ}+\cos 40^{\circ}-\cos 340^{\circ}\right]$

$=\frac{1}{2}\left[-\frac{3}{2}+\left\{2 \cos \left(\frac{80^{\circ}+40^{\circ}}{2}\right) \cos \left(\frac{80^{\circ}-40^{\circ}}{2}\right)-\cos \left(360^{\circ}-20^{\circ}\right)\right\}\right]$

$=\frac{1}{2}\left[-\frac{3}{2}+\left\{2 \cos 60^{\circ} \cos 20^{\circ}-\cos 20^{\circ}\right\}\right]$

$=\frac{1}{2}\left[-\frac{3}{2}+\cos 20^{\circ}-\cos 20^{\circ}\right]$

$=\frac{1}{2}\left[-\frac{3}{2}\right]$

$=-\frac{3}{4}=\mathrm{RHS}$

Hence, LHS = RHS

(vi) Consider LHS :

$\mathrm{LHS}=\sin \frac{x}{2} \sin \frac{7 x}{2}+\sin \frac{3 x}{2} \sin \frac{11 x}{2}$

$=\frac{1}{2}\left[2 \sin \frac{x}{2} \sin \frac{7 x}{2}+2 \sin \frac{3 x}{2} \sin \frac{11 x}{2}\right]$

$=\frac{1}{2}\left[\cos \left(\frac{7 x}{2}-\frac{x}{2}\right)-\cos \left(\frac{7 x}{2}+\frac{x}{2}\right)+\cos \left(\frac{11 x}{2}-\frac{3 x}{2}\right)-\cos \left(\frac{11 x}{2}+\frac{3 x}{2}\right)\right]$

$=\frac{1}{2}[\cos 3 x-\cos 4 x+\cos 4 x-\cos 7 x]$

$=\frac{1}{2}[\cos 3 x-\cos 7 x]$

$=\frac{1}{2}\left[-2 \sin \left(\frac{3 x+7 x}{2}\right) \sin \left(\frac{3 x-7 x}{2}\right)\right]$

$=\frac{1}{2}[-2 \sin (5 x) \sin (-2 x)]$

$=\sin (5 x) \sin (2 x)=\mathrm{RHS}$

Hence, LHS = RHS

(vii) Consider LHS :

LHS $=\cos x \cos \frac{x}{2}-\cos 3 x \cos \frac{9 x}{2}$

$=\frac{1}{2}\left[2 \cos x \cos \frac{x}{2}-2 \cos 3 x \cos \frac{9 x}{2}\right]$

$=\frac{1}{2}\left[\cos \left(x+\frac{x}{2}\right)+\cos \left(x-\frac{x}{2}\right)-\cos \left(3 x+\frac{9 x}{2}\right)-\cos \left(3 x-\frac{9 x}{2}\right)\right]$

$=\frac{1}{2}\left[\cos \frac{3 x}{2}+\cos \frac{x}{2}-\cos \frac{15 x}{2}-\cos \frac{3 x}{2}\right]$

$=\frac{1}{2}\left[\cos \frac{x}{2}-\cos \frac{15 x}{2}\right]$

$=\frac{1}{2}\left[-2 \sin \left(\frac{x+15 x}{4}\right) \sin \left(\frac{x-15 x}{4}\right)\right]$

$=\frac{1}{2}\left[-2 \sin (4 x) \sin \left(-\frac{7 x}{2}\right)\right]$

$=\sin (4 x) \sin \left(\frac{7 x}{2}\right)=\mathrm{RHS}$

Hence, LHS = RHS

Disclaimer: The given question is incorrect. The correct question should be $\cos x \cos \frac{x}{2}-\cos 3 x \cos \frac{9 x}{2}=\sin 4 x \sin \frac{7 x}{2}$.

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