Question

If $f\left(x\right)=co{s}^2\left(x\right)+co{s}^2(x+{120}^0)+co{s}^2(x-{120}^0)$, then value of f $\left(\frac{\pi }{3}\right)$ is-

• A. $1$
• B. $\frac{3}{2}$
• C. $0$
• D. $-1$

Answer 1

The correct option is B $\frac{3}{2}$

We have,

$f\left(x\right)={\cos }^{2}x+{\cos }^2(x+{120}^o)+{\cos }^2(x-{120}^o)$

Then, the value of

$f\left(\frac{\pi }{3}\right)=?$

$f\left(\frac{\pi }{3}\right)={\cos }^2\left(\frac{\pi }{3}\right)+{\cos }^2(\frac{\pi }{3}+{120}^o)+{\cos }^2(\frac{\pi }{3}-{120}^o)$

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

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

$={\cos }^2\frac{\pi }{3}+{\cos }^2\pi +{\cos }^2\frac{\pi }{3}$

$=2{\cos }^2\frac{\pi }{3}+{\cos }^2\pi$

$=2\times {\left(\frac{1}{2}\right)}^2+{(-1)}^2$

$=\frac{1}{2}+1$

$=\frac{3}{2}$

Hence, this is the answer.