Question:

Find the sub-interval of \( (0, \pi/2) \) in which \( f(x) = \log (\sin x + \cos x) \) is increasing and decreasing.

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Use the unit circle or tan graph to quickly identify where \( \tan x < 1 \).
\( \sin x \) and \( \cos x \) intersect at \( x = \pi/4 \) in the first quadrant.
Updated On: Sep 10, 2026
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Solution and Explanation

Concept:

• A function is increasing if \( f'(x) > 0 \) and decreasing if \( f'(x) < 0 \).
• Derivative of \( \log u \) is \( \frac{1}{u} \cdot \frac{du}{dx} \).

Step 1:
Differentiate the function
Given \( f(x) = \log(\sin x + \cos x) \).
\[ f'(x) = \frac{1}{\sin x + \cos x} \cdot \frac{d}{dx}(\sin x + \cos x) \]
\[ f'(x) = \frac{\cos x - \sin x}{\sin x + \cos x} \]

Step 2:
Determine the sign of the derivative
In the interval \( (0, \pi/2) \), \( \sin x + \cos x \) is always positive.
The sign of \( f'(x) \) depends solely on the numerator \( \cos x - \sin x \).
For increasing: \( \cos x - \sin x > 0 \implies \cos x > \sin x \implies \tan x < 1 \).
In \( (0, \pi/2) \), this happens when \( x \in (0, \pi/4) \).
For decreasing: \( \cos x - \sin x 1 \).
In \( (0, \pi/2) \), this happens when \( x \in (\pi/4, \pi/2) \).
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