Question

Let $a=cosx+cos(x+\frac{2\pi }{3})+cos(x+\frac{4\pi }{3})$ and $b=sinx+sin(x+\frac{2\pi }{3})+sin(x+\frac{4\pi }{3})$ then which one of the following holds good ?

• A. $a=2b$
• B. $b=2a$
• C. $a+b=0$
• D. $a+b=1$

Answer 1

The correct option is C $a+b=0$

$a=\cos x+\cos (x+\frac{2\pi }{3})+\cos (x+\frac{4\pi }{3})$

$a=\cos x+2\cos \left(\frac{x+\frac{2\pi }{3}+x+\frac{4\pi }{3}}{2}\right)\cos \left(\frac{x+\frac{2\pi }{3}-x-\frac{4\pi }{3}}{2}\right)$

$a=\cos x+2\cos \left(\frac{2x+\frac{6\pi }{3}}{2}\right)\cos \left(\frac{\frac{-2\pi }{3}}{2}\right)$

$a=\cos x+2\cos (\pi +x)\cos \frac{\pi }{3}$

$a=\cos x-2\cos x\cos \frac{\pi }{3}$

$a=\cos x-2\cos x\times \frac{1}{2}$

$a=\cos x-\cos x=0$

$\therefore a=0$

$b=\sin x+\sin (x+\frac{2\pi }{3})+\sin (x+\frac{4\pi }{3})$

$b=\sin x+2\sin \left(\frac{x+\frac{2\pi }{3}+x+\frac{4\pi }{3}}{2}\right)\cos \left(\frac{x+\frac{2\pi }{3}-x-\frac{4\pi }{3}}{2}\right)$

$b=\sin x+2\sin \left(\frac{2x+\frac{6\pi }{3}}{2}\right)\cos \left(\frac{\frac{-2\pi }{3}}{2}\right)$

$b=\sin x+2\sin (\pi +x)\cos \frac{\pi }{3}$

$b=\sin x-2\sin x\cos \frac{\pi }{3}$

$b=\sin x-2\sin x\times \frac{1}{2}$

$b=\sin x-\sin x=0$

$\therefore b=0$

Hence $a+b=0$