Question

Prove that \(\cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right) = \frac{x}{2}\), \(x \in (0, \pi/4)\).

Hint:

The identities \(1+\sin x = (\cos(x/2)+\sin(x/2))^2\) and \(1-\sin x = (\cos(x/2)-\sin(x/2))^2\) are extremely useful for simplifying expressions involving \(\sqrt{1 \pm \sin x}\). Always check the given interval for \(x\) to correctly determine the signs when taking the square root.

Answer 1

Step 1: Understanding the Concept:
The problem requires simplifying a complex trigonometric expression involving an inverse function. The key is to simplify the argument of the \(\cot^{-1}\) function using trigonometric identities. The identities for \(1 \pm \sin x\) are particularly useful.
Step 2: Key Formula or Approach:
1. Use the identities: \[ 1 + \sin x = \cos^2(x/2) + \sin^2(x/2) + 2\sin(x/2)\cos(x/2) = (\cos(x/2) + \sin(x/2))^2 \] \[ 1 - \sin x = \cos^2(x/2) + \sin^2(x/2) - 2\sin(x/2)\cos(x/2) = (\cos(x/2) - \sin(x/2))^2 \] 2. Substitute these into the expression and simplify the square roots, paying attention to the signs based on the given interval for \(x\).
3. Simplify the resulting fraction inside the \(\cot^{-1}\) function.
4. Use the property \(\cot^{-1}(\cot \theta) = \theta\) for \(\theta\) in the appropriate range.
Step 3: Detailed Explanation:
Let's simplify the terms under the square roots. \[ \sqrt{1+\sin x} = \sqrt{(\cos(x/2) + \sin(x/2))^2} = |\cos(x/2) + \sin(x/2)| \] \[ \sqrt{1-\sin x} = \sqrt{(\cos(x/2) - \sin(x/2))^2} = |\cos(x/2) - \sin(x/2)| \] The problem states that \(x \in (0, \pi/4)\). This means \(x/2 \in (0, \pi/8)\).
In this interval \((0, \pi/8)\), both \(\cos(x/2)\) and \(\sin(x/2)\) are positive. Also, \(\cos(x/2)>\sin(x/2)\).
Therefore, \((\cos(x/2) + \sin(x/2))\) is positive, and \((\cos(x/2) - \sin(x/2))\) is also positive.
So we can remove the absolute value signs: \[ \sqrt{1+\sin x} = \cos(x/2) + \sin(x/2) \] \[ \sqrt{1-\sin x} = \cos(x/2) - \sin(x/2) \] Now, substitute these into the main expression's argument: \[ \frac{(\cos(x/2) + \sin(x/2)) + (\cos(x/2) - \sin(x/2))}{(\cos(x/2) + \sin(x/2)) - (\cos(x/2) - \sin(x/2))} \] Simplify the numerator and the denominator: \[ \frac{2\cos(x/2)}{2\sin(x/2)} = \cot(x/2) \] So the original expression becomes: \[ \cot^{-1}(\cot(x/2)) \] Since \(x/2 \in (0, \pi/8)\), which is within the principal value range of \(\cot^{-1}\) (which is \((0, \pi)\)), we can simplify this to: \[ \cot^{-1}(\cot(x/2)) = \frac{x}{2} \] Step 4: Final Answer:
We have proven that \(\cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right) = \frac{x}{2}\) for \(x \in (0, \pi/4)\).