Question

If $2\cos A=x+\dfrac{1}{x},2\cos B=y+\dfrac{1}{y}$ , show that $2\cos \left( A-B \right)=\dfrac{x}{y}+\dfrac{y}{x}$

Answer 1

Hint: To solve this problem, we have to find the range of $x+\dfrac{1}{x}$ as the range of Cosine function is $\left[ -1,1 \right]$. Assume the minimum value of $x+\dfrac{1}{x}=k$ and simplify to get a quadratic equation. Rearrange the terms to make it as a perfect square. Then, to satisfy the condition, equate $x=\dfrac{k}{2}\text{ and }1-\dfrac{{{k}^{2}}}{4}=0$ , solve them further and we will get the range of $x+\dfrac{1}{x}$ and we can find the values of cos A and cos B and eventually cos (A-B).

Complete step-by-step answer:
Let us start the question by assuming the minimum value of $x+\dfrac{1}{x}$ . Now, we will simplify and get as shown below,
$\begin{align}
  & x+\dfrac{1}{x}=k \\
 & \dfrac{{{x}^{2}}+1}{x}=k\Rightarrow {{x}^{2}}+1=kx \\
 & {{x}^{2}}+1-kx=0 \\
\end{align}$
 Making it as a perfect square gives us
$\begin{align}
  & {{x}^{2}}-kx+\dfrac{{{k}^{2}}}{4}+1-\dfrac{{{k}^{2}}}{4}=0 \\
 & {{\left( x-\dfrac{k}{2} \right)}^{2}}+1-\dfrac{{{k}^{2}}}{4}=0\to \left( 1 \right) \\
\end{align}$
We have to get the values of x and the range. So, we will consider the different cases.
$\begin{align}
  & 1-\dfrac{{{k}^{2}}}{4}=0\Rightarrow \dfrac{{{k}^{2}}}{4}=1\Rightarrow {{k}^{2}}=4 \\
 & k=\pm 2 \\
 & For\text{ }k=+2\text{ } \\
 & x=\dfrac{k}{2}\Rightarrow x=1 \\
 & For\text{ }k=-2\text{ } \\
 & x=\dfrac{k}{2}\Rightarrow x=-1 \\
\end{align}$
$\therefore x+\dfrac{1}{x}\in \left( -\infty ,-2 \right)\cup \left( 2,\infty \right)$
But we have $2\cos A=x+\dfrac{1}{x}$
$\begin{align}
  & 2\cos A\ge 2\text{ or }2\cos A\le -2 \\
 & \Rightarrow \cos A\ge 1\text{ or }\cos A\le -1 \\
\end{align}$But the range of cosA is [-1, 1]. From the given constraints we have
cosA = $\pm 1$
When x = 1, 2cos A = 1 +$\dfrac{1}{1}$$\Rightarrow \cos A=1\Rightarrow A=2n\pi \text{ }n\in N$
When x = -1, 2cos A = -1 +$\dfrac{1}{-1}$$\Rightarrow \cos A=-1\Rightarrow A=\left( 2n-1 \right)\pi \text{ n}\in \text{N}$
Similarly we can write for cos B as the equations are same with a variable change.
When y = 1, 2cos B = 1 +$\dfrac{1}{1}$$\Rightarrow \cos B=1\Rightarrow B=2m\pi \text{ m}\in N$
When y = -1, 2cos B = -1 +$\dfrac{1}{-1}$$\Rightarrow \cos B=-1\Rightarrow B=\left( 2m-1 \right)\pi \text{ m}\in \text{N}$
Consider the case x = 1 and y = 1
$A=2n\pi \text{ , }B=2m\pi$
$2\cos (A-B)=2\cos (2n\pi -2m\pi )=2\cos ((2n-2m)\pi )=2\cos (2k\pi )=2$
L.H.S = 2
$\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{1}{1}+\dfrac{1}{1}=2=R.H.S$
L.H.S = R.H.S
Consider the case x = -1 and y = 1
$A=\left( 2n-1 \right)\pi \text{ , }B=2m\pi$
$2\cos (A-B)=2\cos (\left( 2n-1 \right)\pi -2m\pi )=2\cos ((2n-2m-1)\pi )=2\cos (\left( 2k-1 \right)\pi )=-2=L.H.S$$\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{-1}{1}+\dfrac{1}{-1}=-2=R.H.S$
L.H.S = R.H.S
Similar is the case of x = 1 and y = -1
Consider the case x = -1 and y = -1
$A=\left( 2n-1 \right)\pi \text{ , }B=\left( 2m-1 \right)\pi$
$2\cos (A-B)=2\cos (\left( 2n-1 \right)\pi -\left( 2m-1 \right)\pi )=2\cos ((2n-2m-2)\pi )=2\cos (2k\pi )=2=L.H.S$
$\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{-1}{-1}+\dfrac{-1}{-1}=2=R.H.S$
L.H.S = R.H.S
In every case we got that L.H.S = R.H.S
Hence proved the statement
If $2\cos A=x+\dfrac{1}{x},2\cos B=y+\dfrac{1}{y}$ , then $2\cos \left( A-B \right)=\dfrac{x}{y}+\dfrac{y}{x}$

Note:An alternative approach to find the range of $x+\dfrac{1}{x}$is to apply the property of A.M ≥ H.M for positive numbers. This means that
$\begin{align}
  & \dfrac{x+\dfrac{1}{x}}{2}\ge \sqrt{x\times \dfrac{1}{x}} \\
 & \dfrac{x+\dfrac{1}{x}}{2}\ge 1 \\
 & x+\dfrac{1}{x}\ge 2 \\
\end{align}$
We have to be careful that x and y can also take negative values and prove the statement given in question.