Prove that: 20^ + 100^ + 140^ =0

Question

Prove that: $\cos20^\circ+\cos100^\circ+\cos140^\circ=0$

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Answer

$\text{LHS}=\cos20^\circ+\cos100^\circ+\cos140^\circ$ $\Rightarrow\ (\cos20^\circ+\cos100^\circ)+\cos140^\circ$ $=\ 2\cos\Big(\frac{20^\circ+100^\circ}{2}\Big)\cos\Big(\frac{20^\circ-100^\circ}{2}\Big)+\cos140^\circ$ $\Big[\because\ \cos\text{c}-\cos\text{D}=2\cos\Big(\frac{\text{C}-\text{D}}{2}\Big)\cos\Big(\frac{\text{C}-\text{D}}{2}\Big)\Big]$ $=\ 2\cos60^\circ\cos(-40^\circ)+\cos140^\circ$ $=\ 2\times\frac{1}{2}\cos40+\cos140^\circ$ $\Big[\because\ \cos60^\circ=\frac{1}{2}\Big]$ $=\ \cos40^\circ+\cos(180^\circ-40^\circ)$ $=\ \cos40^\circ-\cos40^\circ$ $=\ 0$ $=\ \text{RHS}$

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