Question

If $x = y\cos \dfrac{{2\pi }}{3} = z\cos \dfrac{{4\pi }}{3}$ , then xy + yz + zx is equal to
A.-1
B.0
C.1
D.2

Answer 1

First, simplify the given equation $x = y\cos \dfrac{{2\pi }}{3} = z\cos \dfrac{{4\pi }}{3}$ by converting the trigonometric functions into their respective values.
Now, when we get a relation between x, y and z, substitute the values in the equation xy + yz + zx, to get the answer.

Complete step-by-step answer:
It is given that, $x = y\cos \dfrac{{2\pi }}{3} = z\cos \dfrac{{4\pi }}{3}$ .
Now, first, let’s simplify $\cos \dfrac{{2\pi }}{3}$ and $\cos \dfrac{{4\pi }}{3}$ .

$$\cos \dfrac{{2\pi }}{3} = \cos \left( {\pi - \dfrac{\pi }{3}} \right) \\\ = - \cos \dfrac{\pi }{3} \\\ = - \dfrac{1}{2} \\\$$

And
$\cos \dfrac{{4\pi }}{3} = \cos \left( {\pi + \dfrac{\pi }{3}} \right) \\\ = - \cos \dfrac{\pi }{3} \\\ = - \dfrac{1}{2} \\\$
Now, substituting $\cos \dfrac{{2\pi }}{3} = - \dfrac{1}{2}$ and $\cos \dfrac{{4\pi }}{3} = - \dfrac{1}{2}$ in $x = y\cos \dfrac{{2\pi }}{3} = z\cos \dfrac{{4\pi }}{3}$ gives
$\therefore x = y\left( { - \dfrac{1}{2}} \right) = z\left( { - \dfrac{1}{2}} \right) \\\ \therefore x = - \dfrac{y}{2} = - \dfrac{z}{2} \\\ \therefore - 2x = y = z \\\$
Thus, we get $y = z = - 2x$ .
We have to find xy + yz + zx.
$\therefore$ xy + yz + zx $= x\left( { - 2x} \right) + \left( { - 2x} \right)\left( { - 2x} \right) + x\left( { - 2x} \right)$ $(y = z = - 2x)$
$= - 2{x^2} + 4{x^2} - 2{x^2}$
= 0
Thus, xy + yz + zx = 0.

Note: Here, $\cos \dfrac{{2\pi }}{3} = - \dfrac{1}{2}$ because $\cos \dfrac{{2\pi }}{3} = \cos \left( {\pi - \dfrac{\pi }{3}} \right)$ lies in quadrant 2 and in quadrant 2, the value of any $\cos \theta$ is negative.
Also, $\cos \dfrac{{4\pi }}{3} = - \dfrac{1}{2}$ because $\cos \dfrac{{4\pi }}{3} = \cos \left( {\pi + \dfrac{\pi }{3}} \right)$ lies in quadrant 3 and in quadrant 3, the value of any $\cos \theta$ is negative.