Question

Find the sub-intervals of \[ (0,\pi) \] in which \[ f(x)=\tan^{-1}(\sin x-\cos x) \] is increasing and decreasing.

Hint:

Whenever you encounter linear combinations of sine and cosine functions like \(a\sin x + b\cos x\), convert them into a single sine or cosine function using the transformation factor \(\sqrt{a^2 + b^2}\) to simplify your inequality analysis.

Answer 1

Concept: To find the intervals where a function is increasing or decreasing, we evaluate the sign of its first derivative, \(f'(x)\):
• A function is strictly increasing in an interval if \(f'(x) > 0\) for all points in that interval.
• A function is strictly decreasing in an interval if \(f'(x) < 0\) for all points in that interval. The derivative of \(\tan^{-1}(u)\) with respect to \(x\) is given by the chain rule: \[ \frac{d}{dx}\left(\tan^{-1}(u)\right) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \] Since the denominator term \((1 + u^2)\) is always strictly positive for any real value of \(u\), the overall sign of the derivative depends entirely on the sign of \(\frac{du}{dx}\).

Step 1: Compute the first derivative \(f'(x)\) using the chain rule.
The given function is: \[ f(x) = \tan^{-1}(\sin x - \cos x) \] Differentiating with respect to \(x\): \[ f'(x) = \frac{1}{1 + (\sin x - \cos x)^2} \cdot \frac{d}{dx}(\sin x - \cos x) \] We know the basic derivatives: \(\frac{d}{dx}(\sin x) = \cos x\) and \(\frac{d}{dx}(\cos x) = -\sin x\). Substituting these values: \[ f'(x) = \frac{\cos x - (-\sin x)}{1 + (\sin x - \cos x)^2} \] \[ f'(x) = \frac{\cos x + \sin x}{1 + (\sin x - \cos x)^2} \]

Step 2: Analyze the conditions for the function to be increasing.
For \(f(x)\) to be increasing, we require \(f'(x) > 0\): \[ \frac{\cos x + \sin x}{1 + (\sin x - \cos x)^2} > 0 \] Since the denominator term \((1 + (\sin x - \cos x)^2) \ge 1\), it is always positive. Therefore, the inequality simplifies to: \[ \cos x + \sin x > 0 \] To solve this within the domain interval \((0, \pi)\), let us divide the expression by \(\sqrt{2}\) to rewrite it as a single trigonometric function: \[ \frac{1}{\sqrt{2}}\cos x + \frac{1}{\sqrt{2}}\sin x > 0 \] Using the angle sum identity \(\sin(x + \frac{\pi}{4}) = \sin x\cos\frac{\pi}{4} + \cos x\sin\frac{\pi}{4}\), this simplifies to: \[ \sin\left(x + \frac{\pi}{4}\right) > 0 \]

Step 3: Solve for the intervals within the domain boundaries.
The given domain is \(x \in (0, \pi)\). Let us find the corresponding range for the shifted angle \(\left(x + \frac{\pi}{4}\right)\) by adding \(\frac{\pi}{4}\) to the domain bounds: \[ \frac{\pi}{4} < x + \frac{\pi}{4} < \pi + \frac{\pi}{4} \quad \Rightarrow \quad \left(x + \frac{\pi}{4}\right) \in \left(\frac{\pi}{4}, \frac{5\pi}{4}\right) \] We know that the sine function is positive in the first and second quadrants, which means when its angle lies between \(0\) and \(\pi\).
Therefore, \(\sin\left(x + \frac{\pi}{4}\right) > 0\) when: \[ \frac{\pi}{4} < x + \frac{\pi}{4} < \pi \] Subtracting \(\frac{\pi}{4}\) from all parts of the inequality to isolate \(x\): \[ 0 < x < \pi - \frac{\pi}{4} \quad \Rightarrow \quad 0 < x < \frac{3\pi}{4} \] Thus, the function is strictly increasing in the sub-interval \(\left(0, \frac{3\pi}{4}\right)\).

Step 4: Determine where the function is decreasing.
The function decreases where the derivative is negative, \(f'(x) < 0\), which occurs when the sine function is negative: \[ \sin\left(x + \frac{\pi}{4}\right) < 0 \] Within our specific angle range, the sine function becomes negative in the third quadrant, which is between \(\pi\) and \(\frac{5\pi}{4}\): \[ \pi < x + \frac{\pi}{4} < \frac{5\pi}{4} \] Subtracting \(\frac{\pi}{4}\) from all parts of the inequality to isolate \(x\): \[ \pi - \frac{\pi}{4} < x < \frac{5\pi}{4} - \frac{\pi}{4} \quad \Rightarrow \quad \frac{3\pi}{4} < x < \pi \] Thus, the function is strictly decreasing in the sub-interval \(\left(\frac{3\pi}{4}, \pi\right)\). This matches option (C).