Question

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

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

(b) 1

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

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

Answer 1

$$\begin{gathered} \cos 12°+\cos 84°+\cos 156°+\cos 132° \\ =\cos 12°+\cos 84°+\cos \left(180°-24\right)+\cos \left(108°-4\mathrm{F}\right) \\ =\cos 12°+\cos 84°-\cos 24°-\cos 48°\left(\because \cos \left(180-\theta \right)=-\cos \theta \right) \\ =\cos 12°-\cos 48°+\cos 84°-\cos 24° \\ =-2\sin \left(\frac{12°+48°}{2}\right)\sin \left(\frac{12°-48°}{2}\right)-2\sin \left(\frac{84°+24°}{2}\right)\sin \left(\frac{84°-24°}{2}\right) \\ \left(\mathrm{using}\cos a-\cos b=2\sin \left(\frac{b-a}{2}\right)\sin \left(\frac{a+b}{2}\right)\right) \\ =2\sin 30°\sin 18°-2\sin 54°\sin 30° \\ =2\sin 30°\left(\sin 18°-\sin 54°\right) \\ =2\times \frac{1}{2}\left[2\sin \left(\frac{18°-54°}{2}\right)\cos \left(\frac{18°+54°}{2}\right)\right]\left[\sin a-\sin b=2\sin \frac{a-b}{2}\cos \frac{a+b}{2}\right] \\ =\left[-2\sin 18°\cos 36°\right] \\ =-2\sin 18°\cos 36°\left(\frac{\cos 18°}{\cos 18°}\right) \\ =\frac{-\left(2\sin 18°\cos 18°\right)\cos 36°}{\cos 18°} \\ =\frac{-\sin 36°\cos 36°}{\cos 18°}=\frac{-2\sin 36°\cos 36°}{2\cos 18°} \\ \mathrm{Hence}\cos 12°+\cos 4°+\cos 156°+\cos 132°=\frac{-\sin 72°}{2\sin \left(90°-72°\right)}=\frac{-\sin 72°}{2\sin 72°}=\frac{-1}{2} \end{gathered}$$