If x = y cos \(\frac{2π}{3}\) = z cos \(\frac{4π}{3}\), then what i

1.	If x = y cos \(\frac{2π}{3}\) = z cos \(\frac{4π}{3}\), then what is xy + yz + zx equal to?
Answer» \(x\) = y cos 120° = z cos 240° ⇒ \(x\) = y cos (180° – 60°) = z cos (180° + 60°) ⇒ \(x\) = – y cos 60° = – z cos 60° (∵ cos (180° – θ) = – cos θ = cos (180° + θ)) ⇒ \(x\) = \(-\frac12\) y = \(-\frac{z}2\) ⇒ 2\(x\) = – y = – z ⇒ \(\frac{x}{\frac12}\) = \(\frac{y}{(-1)}\) = \(\frac{z}{(-1)}\) = k ⇒ \(x\) = \(\frac{k}{2}\) , y = – k, z = – k ∴ xy + yz + zx = \(\big(\frac{k}{2}\big)\) (–k) + (–k) (–k) + (–k) \(\big(\frac{k}{2}\big)\) = \(\frac{-k^2}{2}\) + k² – \(\frac{k^2}{2}\) = 0.

If x = y cos \(\frac{2π}{3}\) = z cos \(\frac{4π}{3}\), then what is xy + yz + zx equal to?

Answer»

\(x\) = y cos 120° = z cos 240°

⇒ \(x\) = y cos (180° – 60°) = z cos (180° + 60°)

⇒ \(x\) = – y cos 60° = – z cos 60° (∵ cos (180° – θ) = – cos θ = cos (180° + θ))

⇒ \(x\) = \(-\frac12\) y = \(-\frac{z}2\) ⇒ 2\(x\) = – y = – z

⇒ \(\frac{x}{\frac12}\) = \(\frac{y}{(-1)}\) = \(\frac{z}{(-1)}\) = k

⇒ \(x\) = \(\frac{k}{2}\) , y = – k, z = – k

∴ xy + yz + zx = \(\big(\frac{k}{2}\big)\) (–k) + (–k) (–k) + (–k) \(\big(\frac{k}{2}\big)\) = \(\frac{-k^2}{2}\) + k² – \(\frac{k^2}{2}\) = 0.