Question

Then the value of $x+y+z$ is equal to :

• A. $0$
• B. $1$
• C. $p+q+r$
• D. None of these

Answer 1

Answer: (A), (C)

$0$
$p+q+r$

$\alpha ,\beta ,\gamma$ are in A.P. with common difference $\frac{2\pi }{3}$

The angles will be $\alpha$, $\alpha +\frac{2\pi }{3}$, $\alpha +\frac{4\pi }{3}$

Now,

$\sin \left(\alpha \right)+\sin (\alpha +\frac{2\pi }{3})+\sin (\alpha +\frac{4\pi }{3})$

$=\sin \left(\alpha \right)+\sin (\pi -(\frac{\pi }{3}-\alpha ))+\sin (\pi +\alpha +\frac{\pi }{3})$

$=\sin \left(\alpha \right)-\sin (\alpha +\frac{\pi }{3})+\sin (\frac{\pi }{3}-\alpha )$

$=\sin \left(\alpha \right)-\sin (\alpha +\frac{\pi }{3})-\sin (\alpha -\frac{\pi }{3})$

$=\sin \left(\alpha \right)-[\sin (\alpha +\frac{\pi }{3})+\sin (\alpha -\frac{\pi }{3})]$

$=\sin \left(\alpha \right)-\left[2\sin \left(\alpha \right)\cos \left(\frac{\pi }{3}\right)\right]$

$=\sin \alpha -\sin \alpha$

$=0$

Hence

$x+y+z=0$ ...$\left(i\right)$

Similarly

$\cos \left(\alpha \right)+\cos (\alpha +\frac{2\pi }{3})+\cos (\alpha +\frac{4\pi }{3})$

$=\cos \left(\alpha \right)+\cos (\pi -(\frac{\pi }{3}-\alpha ))+\cos (\pi +\alpha +\frac{\pi }{3})$

$=\cos \left(\alpha \right)-\cos (\alpha +\frac{\pi }{3})-\cos (\frac{\pi }{3}-\alpha )$

$=\cos \left(\alpha \right)-\cos (\alpha +\frac{\pi }{3})-\cos (\alpha -\frac{\pi }{3})$

$=\cos \left(\alpha \right)-[\cos (\alpha +\frac{\pi }{3})+\cos (\alpha -\frac{\pi }{3})]$

$=\cos \left(\alpha \right)-\left[2\cos \left(\alpha \right)\cos \left(\frac{\pi }{3}\right)\right]$

$=0$

$=p+q+r$.

Hence, $x+y+z=0=p+q+r$.