The solution of the equation `1-"cos" theta = "sin" theta "sin" (theta)/(2)` is

Options

A. `n pi, n in Z`
B. `2n pi, n in Z`
C. `(npi)/(2), n in Z`
D. none of these

Answer

To solve the equation \( 1 - \cos \theta = \sin \theta \cdot \frac{\sin \theta}{2} \), we can follow these steps:

### Step 1: Rewrite the Equation
Start with the given equation:
\[ 1 - \cos \theta = \sin \theta \cdot \frac{\sin \theta}{2} \]
This simplifies to:
\[ 1 - \cos \theta = \frac{1}{2} \sin^2 \theta \]

### Step 2: Use Trigonometric Identities
We can use the identity \( \cos 2\theta = 1 - 2\sin^2 \theta \) to express \( 1 - \cos \theta \) in terms of sine. We know that:
\[ 1 - \cos \theta = 2\sin^2\left(\frac{\theta}{2}\right) \]
Thus, we can rewrite the equation as:
\[ 2\sin^2\left(\frac{\theta}{2}\right) = \frac{1}{2} \sin^2 \theta \]

### Step 3: Substitute for \(\sin^2 \theta\)
Using the identity \( \sin \theta = 2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right) \), we have:
\[ \sin^2 \theta = 4 \sin^2\left(\frac{\theta}{2}\right) \cos^2\left(\frac{\theta}{2}\right) \]
Substituting this into our equation gives:
\[ 2\sin^2\left(\frac{\theta}{2}\right) = \frac{1}{2} \cdot 4 \sin^2\left(\frac{\theta}{2}\right) \cos^2\left(\frac{\theta}{2}\right) \]
This simplifies to:
\[ 2\sin^2\left(\frac{\theta}{2}\right) = 2 \sin^2\left(\frac{\theta}{2}\right) \cos^2\left(\frac{\theta}{2}\right) \]

### Step 4: Factor the Equation
Rearranging gives:
\[ 2\sin^2\left(\frac{\theta}{2}\right) - 2 \sin^2\left(\frac{\theta}{2}\right) \cos^2\left(\frac{\theta}{2}\right) = 0 \]
Factoring out \( 2\sin^2\left(\frac{\theta}{2}\right) \):
\[ 2\sin^2\left(\frac{\theta}{2}\right) (1 - \cos^2\left(\frac{\theta}{2}\right)) = 0 \]
Using the identity \( 1 - \cos^2 x = \sin^2 x \), we have:
\[ 2\sin^2\left(\frac{\theta}{2}\right) \sin^2\left(\frac{\theta}{2}\right) = 0 \]

### Step 5: Solve for \(\theta\)
This gives us two cases:
1. \( \sin^2\left(\frac{\theta}{2}\right) = 0 \)
2. \( \sin^2\left(\frac{\theta}{2}\right) = 0 \) (which is the same case)

From \( \sin\left(\frac{\theta}{2}\right) = 0 \), we find:
\[ \frac{\theta}{2} = n\pi \quad \Rightarrow \quad \theta = 2n\pi \]
where \( n \) is any integer.

### Final Solution
Thus, the solution to the equation is:
\[ \theta = 2n\pi, \quad n \in \mathbb{Z} \]