Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

We have $cos\frac{2\pi }{3}=cos\frac{2\times {180}^0}{3}=cos{120}^0=cos({180}^2-{60}^0)=-cos{60}^0=-\frac{1}{2}$

Also $cos\frac{4\pi }{3}=cos\frac{4\times {180}^0}{3}=cos{240}^0=cos({180}^2+{60}^0)=-cos{60}^0=-\frac{1}{2}$

So, $x=y\times (-\frac{1}{2})=z\times (-\frac{1}{2})$
$=>y=-2x;z=-2x$

$=>xy+yz+zx=x(-2x)+(-2x)(-2x)+(-2x)x=-2x^2+4x^2-2x^2=0$