The value of cos 12° + cos 84° + cos 156° + cos 132° is
Question:

The value of cos 12° + cos 84° + cos 156° + cos 132° is

(a) $\frac{1}{2}$

(b) 1

(C) $-\frac{1}{2}$

 

(d) $-\frac{1}{8}$

Solution:

$\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}$

$=\cos 12^{\circ}+\cos 84^{\circ}+\cos \left(180^{\circ}-24\right)+\cos \left(108^{\circ}-4 \mathrm{~F}\right)$

$=\cos 12^{\circ}+\cos 84^{\circ}-\cos 24^{\circ}-\cos 48^{\circ}(\because \cos (180-\theta)=-\cos \theta)$

$=\cos 12^{\circ}-\cos 48^{\circ}+\cos 84^{\circ}-\cos 24^{\circ}$

$=-2 \sin \left(\frac{12^{\circ}+48^{\circ}}{2}\right) \sin \left(\frac{12^{\circ}-48^{\circ}}{2}\right)-2 \sin \left(\frac{84^{\circ}+24^{\circ}}{2}\right) \sin \left(\frac{84^{\circ}-24^{\circ}}{2}\right)$

$\left(\right.$ using $\left.\cos a-\cos b=2 \sin \left(\frac{b-a}{2}\right) \sin \left(\frac{a+b}{2}\right)\right)$

$=2 \sin 30^{\circ} \sin 18^{\circ}-2 \sin 54^{\circ} \sin 30^{\circ}$

$=2 \sin 30^{\circ}\left(\sin 18^{\circ}-\sin 54^{\circ}\right)$

$=2 \times \frac{1}{2}\left[2 \sin \left(\frac{18^{\circ}-54^{\circ}}{2}\right) \cos \left(\frac{18^{\circ}+54^{\circ}}{2}\right)\right] \quad\left[\sin a-\sin b=2 \sin \frac{a-b}{2} \cos \frac{a+b}{2}\right]$

$=\left[-2 \sin 18^{\circ} \cos 36^{\circ}\right]$

$=-2 \sin 18^{\circ} \cos 36^{\circ}\left(\frac{\cos 18^{\circ}}{\cos 18^{\circ}}\right)$

$=\frac{-\left(2 \sin 18^{\circ} \cos 18^{\circ}\right) \cos 36^{\circ}}{\cos 18^{\circ}}$

$=\frac{-\sin 36^{\circ} \cos 36^{\circ}}{\cos 18^{\circ}}=\frac{-2 \sin 36^{\circ} \cos 36^{\circ}}{2 \cos 18^{\circ}}$

Hence $\cos 12^{\circ}+\cos 4^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}=\frac{-\sin 72^{\circ}}{2 \sin \left(90^{\circ}-72^{\circ}\right)}=\frac{-\sin 72^{\circ}}{2 \sin 72^{\circ}}=\frac{-1}{2}$

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