Question

For $x,y,z\in \left(0,\frac{\pi }{2}\right)$, let $x,y,z$ be first three consecutive terms of an arithmetic progression such that $\cos x+\cos y+\cos z=1$ and $\sin x+\sin y+\sin z=\frac{1}{\sqrt{2}}$, then which of the following is/ are correct?

• A. $\cot y=\sqrt{2}$
• B. $\cos (x-y)=\frac{\sqrt{3}-1}{2\sqrt{2}}$
• C. $\tan 2y=\frac{2\sqrt{2}}{3}$
• D. $\sin (x-y)+\sin (z-y)=0$

Answer 1

Answer: (A), (D)

We have $y_{{\{}_{=x}}-d,y,y_{{\{}_{=z}}+d$ in A.P
Now, $\sum \cos x=1\Rightarrow \cos (y-d)+\cos y+\cos (y+d)=1\Rightarrow \cos y(2\cos d+1)=1$ ....(1)
Also, $\sum \sin x=\frac{1}{\sqrt{2}}\Rightarrow \sin (y-d)+\sin y+\sin (y+d)=\frac{1}{\sqrt{2}}\Rightarrow \sin y(2\cos d+1)=\frac{1}{\sqrt{2}}$ ....(2)
$\therefore \frac{\text{ Equation (1) }}{\text{ Equation (2) }}\Rightarrow \cot y=\sqrt{2}$

Now, putting $\cos y=\frac{\sqrt{2}}{\sqrt{3}}$ in (1), we get
$2\cos d+1=\frac{\sqrt{2}}{\sqrt{3}}\Rightarrow \cos d=\frac{\sqrt{3}-\sqrt{2}}{2\sqrt{2}}=\cos (y-x)=\cos (x-y)$
Also, $\tan 2y=\frac{2\tan y}{1-{\tan }^{2}y}=\frac{2\times \frac{1}{\sqrt{2}}}{1-\frac{1}{2}}=\frac{\sqrt{2}}{\frac{1}{2}}=2\sqrt{2}$
Clearly, $\sin (x-y)+\sin (z-y)=\sin (-d)+\sin d=0$.