Question

If $x$ and $y$ are acute angles such that $\cos x=\frac{13}{14}\text{and}\cos y=\frac{1}{7}$ prove that $\left(x-y\right)=-\frac{\pi }{3}$,

Answer 1

given, $\cos x=\frac{13}{14}\cos y=\frac{1}{y}$

$\because x$ & $y$ are acute angles

$\therefore \sin x=\sqrt{1-{\cos }^{2}x}=\sqrt{1-\frac{169}{196}}=\sqrt{\frac{27}{196}}$

$\sin x=\frac{3\sqrt{3}}{14}$and $\sin y=\sqrt{1-{\cos }^{2}y}=\sqrt{1-\frac{1}{49}}=\sqrt{\frac{48}{49}}$

$\sin y=\frac{4\sqrt{3}}{7}$

Now $\cos (x-y)=\csc x\cos y+\sin x\sin y$

$=\frac{13}{14}\times \frac{1}{7}+\frac{3\sqrt{3}}{14}\times \frac{4\sqrt{3}}{7}$

$=\frac{13+36}{14\times 7}$

$=\frac{49}{14\times 7}$

$=\frac{1}{2}$

$\therefore |x-y|={\cos }^{-1}\frac{1}{2}=\frac{\pi }{3}$

$\Rightarrow |x-y|=\frac{\pi }{3}-----\left(1\right)$

Now, we know $\cos \theta$ is a decreasing function

and $\frac{13}{14}>\frac{1}{7}$

$\Rightarrow \cos x>\cos y\Rightarrow x<y$

$\Rightarrow x-y<0$

from $\left(1\right)$

$\therefore -(x-y)=\frac{\pi }{3}\Rightarrow x-y=\frac{-\pi }{3}$ Proved