Question

2 cos A = x + $\frac{1}{x}$ , 2cosB = y + $\frac{1}{y}$ . Find 2 cos(A-B)

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

To find cos (A - B), we need cosA, cosB, sinA and sinB. Since we know cosA, we can find sinA, similarly sinB also. Then substitute in the formula cos (A-B) = cosAcosB+sinAsinB. This is straight forward.

We will try to approach the problem with the help of complex numbers. $\frac{1}{2}(x+\frac{1}{x})$ Is the A.M of x and $\frac{1}{x}$ It should be atleast their G.M.

$\Rightarrow$ $\frac{1}{2}(x+\frac{1}{x})\ge \sqrt{X\frac{1}{x}}$ = 1

$\frac{1}{2}(x+\frac{1}{x})=cosA\ge 1$

⇒cosA should be equal to 1, because it can't be more than 1.

This means x =1 is the only real value which satisfies it. This motivates us to treat x as a complex number, because we are not given x is a real number.

We know e$i^{\theta }+e^{-i\theta }=2cos\theta$

⇒We can take x = eiA and y = eiB $[\because 2cosA=e^{iA}+\frac{1}{e^{iA}}]$

We want to find 2 cos (A-B)

⇒ 2 cos (A-B) = $e^{i(A-B)}+e^{-1(A-B)}$

= $\frac{e^{iA}}{e^{iB}}+\frac{e^{iB}}{e^{iA}}$

= $\frac{x}{y}+\frac{y}{x}$

Key steps: (1) Find sinA, sinB and using the formula of cos (A-B)

Or (2) Treating x and y as complex numbers

(3) $e^{i\theta }+e^{-1\theta }$ = 2cos $\theta$