For `-pi/2 lt theta lt pi/2, (sintheta + sin 2theta)/(1+costheta + cos2theta)` lies in the interval

Options

A. `(-infty, infty)`
B. `(-2,-2)`
C. `(0,infty)`
D. `(-1,1)`

Answer

To solve the expression \((\sin \theta + \sin 2\theta)/(1 + \cos \theta + \cos 2\theta)\) for \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\), we will simplify the expression step by step.

### Step 1: Rewrite the expression using trigonometric identities
We know that:
- \(\sin 2\theta = 2 \sin \theta \cos \theta\)
- \(\cos 2\theta = 2 \cos^2 \theta - 1\)

Substituting these identities into the expression gives:
\[ \frac{\sin \theta + 2 \sin \theta \cos \theta}{1 + \cos \theta + (2 \cos^2 \theta - 1)} \]

### Step 2: Simplify the denominator
The denominator simplifies as follows:
\[ 1 + \cos \theta + 2 \cos^2 \theta - 1 = \cos \theta + 2 \cos^2 \theta \]
Thus, our expression now looks like:
\[ \frac{\sin \theta + 2 \sin \theta \cos \theta}{\cos \theta + 2 \cos^2 \theta} \]

### Step 3: Factor out common terms
We can factor out \(\sin \theta\) from the numerator:
\[ \frac{\sin \theta(1 + 2 \cos \theta)}{\cos \theta(1 + 2 \cos \theta)} \]

### Step 4: Cancel common factors
Assuming \(1 + 2 \cos \theta \neq 0\), we can cancel \(1 + 2 \cos \theta\) from the numerator and denominator:
\[ \frac{\sin \theta}{\cos \theta} = \tan \theta \]

### Step 5: Determine the range of \(\tan \theta\)
For \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\), the function \(\tan \theta\) is defined and continuous. The range of \(\tan \theta\) in this interval is:
\[ (-\infty, +\infty) \]

### Conclusion
Thus, the expression \(\frac{\sin \theta + \sin 2\theta}{1 + \cos \theta + \cos 2\theta}\) lies in the interval \((- \infty, + \infty)\).