The range of values of `theta in [0, 2pi]` for which `(1+ cos theta, sin theta)` is on interior point of the circle `x^(2) +y^(2)=1`, is

Options

A. `(pi//6, 5pi//6)`
B. `(2pu//3, 5 pi //3)`
C. `(pi//6, 7 pi//6)`
D. `(2pi//3, 4pi//3)`

Answer

To find the range of values of \( \theta \) in the interval \([0, 2\pi]\) for which the point \((1 + \cos \theta, \sin \theta)\) is an interior point of the circle defined by the equation \(x^2 + y^2 = 1\), we will follow these steps:

### Step 1: Set up the inequality
The point \((1 + \cos \theta, \sin \theta)\) will be an interior point of the circle \(x^2 + y^2 < 1\). Therefore, we need to set up the inequality:
\[ (1 + \cos \theta)^2 + (\sin \theta)^2 < 1 \]

### Step 2: Expand the left-hand side
Expanding the left-hand side of the inequality:
\[ (1 + \cos \theta)^2 + \sin^2 \theta = 1 + 2\cos \theta + \cos^2 \theta + \sin^2 \theta \]
Using the Pythagorean identity \(\cos^2 \theta + \sin^2 \theta = 1\), we can simplify this to:
\[ 1 + 2\cos \theta + 1 = 2 + 2\cos \theta \]

### Step 3: Rewrite the inequality
Now we rewrite the inequality:
\[ 2 + 2\cos \theta < 1 \]
Subtracting 2 from both sides gives:
\[ 2\cos \theta < -1 \]

### Step 4: Divide by 2
Dividing both sides by 2, we have:
\[ \cos \theta < -\frac{1}{2} \]

### Step 5: Determine the values of \( \theta \)
Now, we need to find the values of \( \theta \) in the interval \([0, 2\pi]\) for which \(\cos \theta < -\frac{1}{2}\). The cosine function is less than \(-\frac{1}{2}\) in the intervals:
\[ \theta \in \left(\frac{2\pi}{3}, \frac{4\pi}{3}\right) \]

### Final Answer
Thus, the range of values of \( \theta \) for which \((1 + \cos \theta, \sin \theta)\) is an interior point of the circle \(x^2 + y^2 = 1\) is:
\[ \theta \in \left(\frac{2\pi}{3}, \frac{4\pi}{3}\right) \]