Question

Prove that the function \[ f(x) = \tan^{-1} (\sin x + \cos x), x > 0 \text{ is always increasing function on } (0, \frac{\pi}{4}). \]

Hint:

To prove that a function is increasing, check if its derivative is positive on the interval. If the derivative is positive, the function is increasing.

Answer 1

We are given the function \[ f(x) = \tan^{-1} (\sin x + \cos x). \] To prove that \( f(x) \) is increasing on \( \left( 0, \frac{\pi}{4} \right) \), we need to show that the derivative \( f'(x) > 0 \) for \( x \in \left( 0, \frac{\pi}{4} \right) \). First, differentiate \( f(x) \) using the chain rule: \[ f'(x) = \frac{d}{dx} \left( \tan^{-1} (\sin x + \cos x) \right). \] The derivative of \( \tan^{-1} u \) with respect to \( u \) is \( \frac{1}{1+u^2} \). So, applying the chain rule: \[ f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot \frac{d}{dx} (\sin x + \cos x). \] Now, differentiate \( \sin x + \cos x \): \[ \frac{d}{dx} (\sin x + \cos x) = \cos x - \sin x. \] Thus, \[ f'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2}. \] Now, let's analyze the sign of \( f'(x) \) on \( \left( 0, \frac{\pi}{4} \right) \). Note that: - For \( x \in \left( 0, \frac{\pi}{4} \right) \), both \( \cos x \) and \( \sin x \) are positive. - Since \( \cos x > \sin x \) for \( x \in \left( 0, \frac{\pi}{4} \right) \), the numerator \( \cos x - \sin x \) is positive. - The denominator \( 1 + (\sin x + \cos x)^2 \) is always positive because the square of any real number is non-negative, and adding 1 ensures that it is strictly positive. Therefore, \( f'(x) > 0 \) for \( x \in \left( 0, \frac{\pi}{4} \right) \), which means that \( f(x) \) is an increasing function on \( \left( 0, \frac{\pi}{4} \right) \). Conclusion: Since \( f'(x) > 0 \) for \( x \in \left( 0, \frac{\pi}{4} \right) \), the function \( f(x) = \tan^{-1} (\sin x + \cos x) \) is always increasing on \( \left( 0, \frac{\pi}{4} \right) \).