QPaperGen

We are given:
\[
y = \frac{4\sin \theta}{2 + \cos \theta} - \theta
\]

Differentiate with respect to \( \theta \):
\[
\frac{dy}{d\theta} = \frac{(2 + \cos \theta)(4\cos \theta) - 4\sin \theta(-\sin \theta)}{(2 + \cos \theta)^2} - 1
\]

\[
= \frac{8\cos \theta + 4\cos^2 \theta + 4\sin^2 \theta}{(2 + \cos \theta)^2} - 1
\]

Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[
= \frac{8\cos \theta + 4(\cos^2 \theta + \sin^2 \theta)}{(2 + \cos \theta)^2} - 1 = \frac{8\cos \theta + 4}{(2 + \cos \theta)^2} - 1
\]

Now, set the derivative equal to zero:
\[
\frac{dy}{d\theta} = 0 \Rightarrow \frac{8\cos \theta + 4}{(2 + \cos \theta)^2} = 1
\]

Multiply both sides by \( (2 + \cos \theta)^2 \):
\[
8\cos \theta + 4 = (2 + \cos \theta)^2 = 4 + 4\cos \theta + \cos^2 \theta
\]

Simplify:
\[
8\cos \theta + 4 = 4 + 4\cos \theta + \cos^2 \theta \Rightarrow 4\cos \theta = \cos^2 \theta
\]

\[
\Rightarrow \cos \theta (4 - \cos \theta) = 0
\]

So, \( \cos \theta = 0 \) or \( \cos \theta = 4 \). Since \( \cos \theta \in [0,1] \), the only valid solution is \( \cos \theta = 0 \Rightarrow \theta = \frac{\pi}{2} \).

Now express the derivative in a factored form:
\[
\frac{dy}{d\theta} = \frac{8\cos \theta + 4 - (4 + 4\cos \theta + \cos^2 \theta)}{(2 + \cos \theta)^2} = \frac{4\cos \theta - \cos^2 \theta}{(2 + \cos \theta)^2}
\]

\[
= \frac{\cos \theta(4 - \cos \theta)}{(2 + \cos \theta)^2}
\]

Now analyze the sign of \( \frac{dy}{d\theta} \) in \( \left[0, \frac{\pi}{2} \right] \):
\[\]

In this interval, \( \cos \theta > 0 \), and \( 4 - \cos \theta > 0 \) since \( \cos \theta \le 1 \)
\[\] Also, \( (2 + \cos \theta)^2 > 0 \)

So,
\[
\frac{dy}{d\theta} = \frac{\cos \theta (4 - \cos \theta)}{(2 + \cos \theta)^2} > 0 \quad \text{for all } \theta \in \left(0, \frac{\pi}{2} \right)
\]

The derivative is positive throughout the open interval, and since the function is continuous on the closed interval \( \left[0, \frac{\pi}{2} \right] \), it is strictly increasing on the entire interval.