Question

If $x=y\cos \frac{2\pi }{3}=z\cos \frac{4\pi }{3},$ then $xy+yz+zx=$

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (B)

0

$\cos \frac{2\pi }{3}=\cos (\pi -\frac{\pi }{3})=-\cos \frac{\pi }{3}=\frac{-1}{2}$

$\cos \frac{4\pi }{3}=\cos (\pi +\frac{\pi }{3})=-\cos \frac{\pi }{3}=-1/2$

Given $x=y\cos \frac{2\pi }{3}=z\cos \frac{4\pi }{3}$

$\Rightarrow x=y(-1/2)=z(-1/2)$

$\Rightarrow x=y(-1/2)=z(-1/2)$

$\Rightarrow x=\frac{-y}{2}=\frac{-z}{2}$

$\Rightarrow y=-2x=z$

Now, $xy+yz+zx$

$\Rightarrow x(-2x)+(-2x)(-2x)+(-2x)\left(x\right)$

$\Rightarrow -2x^2+{4x}^2-2x^2$

$=0$