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

Hint: 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.