Question

The value of $\cos \left(12°\right)+\cos \left(84°\right)+\cos \left(156°\right)+\cos \left(132°\right)$

• A. $\frac{1}{2}$
• B. $1$
• C. $-\frac{1}{2}$
• D. $\frac{1}{8}$

Answer 1

The correct option is C

$-\frac{1}{2}$

Explanation for correct option

Given trigonometric function is $\cos \left(12°\right)+\cos \left(84°\right)+\cos \left(156°\right)+\cos \left(132°\right)$

$$\begin{gathered} \therefore \cos \left(12°\right)+\cos \left(84°\right)+\cos \left(156°\right)+\cos \left(132°\right) \\ =\left[\cos \left(12°\right)+\cos \left(132°\right)\right]+\left[\cos \left(84°\right)+\cos \left(156°\right)\right] \\ =2\cos \left(\frac{132°+12°}{2}\right)\cos \left(\frac{132°-12°}{2}\right)+2\cos \left(\frac{156°+84°}{2}\right)\cos \left(\frac{156°-84°}{2}\right)\left[\because \cos \left(A\right)+\cos \left(B\right)=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)\right] \\ =2\cos \left(72°\right)\cos \left(60°\right)+2\cos \left(120°\right)\cos \left(36°\right) \\ =2\cos \left(72°\right)·\frac{1}{2}+2\left(-\frac{1}{2}\right)\cos \left(36°\right)\left[\because \cos \left(60°\right)=\frac{1}{2},\cos \left(120°\right)=-\frac{1}{2}\right] \\ =\cos \left(72°\right)-\cos \left(36°\right) \\ =\frac{\sqrt{5}-1}{4}-\frac{\sqrt{5}+1}{4}\left[\because \cos \left(36°\right)=\frac{\sqrt{5}+1}{4},\cos \left(72°\right)=\frac{\sqrt{5}-1}{4}\right] \\ =-\frac{1}{2} \end{gathered}$$Hence, the correct option is $\mathrm{Option}\left(\mathrm{C}\right)$