Question

The function $f(\theta)=\sin \theta+\cos \theta,\ 0\leq \theta \leq 2\pi$ is decreasing in the interval:

• A. \(0\leq \theta < \pi\)
• B. \(0\leq \theta \leq \frac{\pi}{4}\)
• C. \(\frac{\pi}{4}\leq \theta \leq \frac{5\pi}{4}\)
• D. \(\frac{5\pi}{4}\leq \theta \leq 2\pi\)
• E. \(\frac{\pi}{2}\leq \theta \leq \frac{3\pi}{2}\)

Hint:

To find where a trigonometric function is increasing or decreasing, always differentiate first and then solve the sign of the derivative carefully over the given interval.

Answer 1

The Correct Option is C

Step 1: Write the given function.
We are given:
\[ f(\theta)=\sin\theta+\cos\theta, \qquad 0\leq \theta \leq 2\pi \] We need to find the interval where this function is decreasing.

Step 2: Differentiate the function.
The derivative is:
\[ f'(\theta)=\cos\theta-\sin\theta \]

Step 3: Use the condition for decreasing function.
A function is decreasing where its derivative is negative. So we require:
\[ \cos\theta-\sin\theta<0 \] which means:
\[ \cos\theta<\sin\theta \]

Step 4: Rewrite the inequality in a standard form.
We can write: \[ \cos\theta-\sin\theta=\sqrt{2}\cos\left(\theta+\frac{\pi}{4}\right) \] So the condition becomes:
\[ \sqrt{2}\cos\left(\theta+\frac{\pi}{4}\right)<0 \] or simply,
\[ \cos\left(\theta+\frac{\pi}{4}\right)<0 \]

Step 5: Find where cosine is negative.
We know that \(\cos \phi<0\) when:
\[ \frac{\pi}{2}<\phi<\frac{3\pi}{2} \] Here, \[ \phi=\theta+\frac{\pi}{4} \] So, \[ \frac{\pi}{2}<\theta+\frac{\pi}{4}<\frac{3\pi}{2} \]

Step 6: Solve for \(\theta\).
Subtract \(\frac{\pi}{4}\) throughout:
\[ \frac{\pi}{4}<\theta<\frac{5\pi}{4} \] Thus, the function is decreasing on:
\[ \left(\frac{\pi}{4},\frac{5\pi}{4}\right) \] In the option form given, this corresponds to:
\[ \frac{\pi}{4}\leq \theta \leq \frac{5\pi}{4} \]

Step 7: State the final answer.
Hence, the function is decreasing in the interval:
\[ \boxed{\frac{\pi}{4}\leq \theta \leq \frac{5\pi}{4}} \] which matches option \((3)\).