Question

If $2cosA=x+\frac{1}{x},2cos\beta =y+\frac{1}{y}$ then show that 2 cos (A-B)=$\frac{x}{y}+\frac{y}{x}$.

Answer 1

Given,

$2cosA=x+\frac{1}{x}$ and $2cosB=y+\frac{1}{y}$

$\therefore$ $sinA=\sqrt{1-{\cos }^{2}A}=\sqrt{1-\frac{1}{4}{(x+\frac{1}{x})}^2}$

$\Rightarrow sinA=\sqrt{1-\frac{1}{4}(x^2+\frac{1}{x^2}+2)}=\sqrt{-\frac{1}{4}(x^2+\frac{1}{x^2})+\frac{1}{2}}=\sqrt{-\frac{1}{4}(x^2+\frac{1}{x^2}-2)}$

$\Rightarrow sinA=\sqrt{-\frac{1}{4}{(x-\frac{1}{2})}^2}=\frac{\sqrt{-1}}{2}(x-\frac{1}{x})=\frac{i}{2}(x-\frac{1}{x})$ ($\because$ $\sqrt{-1}=i$)

$\Rightarrow 2sinA=i(x-\frac{1}{x})$

Similarly, $2sinB=i(y-\frac{1}{y})$

Now,

$\cos (A-B)=cosA.cosB+sinA.sinB$

$\Rightarrow \cos (A-B)=\frac{1}{4}(x+\frac{1}{x})(y+\frac{1}{y})+\frac{i^2}{4}(x-\frac{1}{x})(y-\frac{1}{y})$

$\Rightarrow \cos (A-B)=\frac{1}{4}(xy+\frac{1}{xy}+\frac{x}{y}+\frac{y}{x})-\frac{1}{4}(xy+\frac{1}{xy}-\frac{x}{y}-\frac{y}{x})$

$\Rightarrow \cos (A-B)=\frac{1}{4}[xy+\frac{1}{xy}+\frac{x}{y}+\frac{y}{x}-xy-\frac{1}{xy}+\frac{x}{y}+\frac{y}{x}]$

$\therefore$ $2\cos (A-B)=\frac{x}{y}+\frac{y}{x}$

Hence proved.