Find sum of `sin2alpha+sin3alpha+.......+sinnalpha` where `(n+2)alpha = 2 pi`

Answer

To find the sum \( S = \sin 2\alpha + \sin 3\alpha + \sin 4\alpha + \ldots + \sin n\alpha \) given that \( (n + 2)\alpha = 2\pi \), we can follow these steps:

### Step 1: Express \(\alpha\) in terms of \(n\)
From the equation \( (n + 2)\alpha = 2\pi \), we can solve for \(\alpha\):
\[ \alpha = \frac{2\pi}{n + 2} \]

### Step 2: Substitute \(\alpha\) into the sum
Now substitute \(\alpha\) back into the sum \( S \):
\[ S = \sin\left(2 \cdot \frac{2\pi}{n + 2}\right) + \sin\left(3 \cdot \frac{2\pi}{n + 2}\right) + \sin\left(4 \cdot \frac{2\pi}{n + 2}\right) + \ldots + \sin\left(n \cdot \frac{2\pi}{n + 2}\right) \]
This simplifies to:
\[ S = \sin\left(\frac{4\pi}{n + 2}\right) + \sin\left(\frac{6\pi}{n + 2}\right) + \sin\left(\frac{8\pi}{n + 2}\right) + \ldots + \sin\left(\frac{2n\pi}{n + 2}\right) \]

### Step 3: Recognize the pattern
The terms can be rewritten as:
\[ S = \sum_{k=2}^{n} \sin\left(\frac{2k\pi}{n + 2}\right) \]

### Step 4: Use the identity for the sum of sine functions
We can use the identity:
\[ \sum_{k=0}^{m-1} \sin\left(a + kd\right) = \frac{\sin\left(\frac{md}{2}\right) \sin\left(a + \frac{(m-1)d}{2}\right)}{\sin\left(\frac{d}{2}\right)} \]
In our case, \( a = \frac{4\pi}{n + 2} \), \( d = \frac{2\pi}{n + 2} \), and \( m = n - 1 \).

### Step 5: Apply the identity
Using the identity:
\[ S = \frac{\sin\left(\frac{(n-1)\cdot 2\pi}{2(n + 2)}\right) \sin\left(\frac{4\pi}{n + 2} + \frac{(n-2)\cdot 2\pi}{2(n + 2)}\right)}{\sin\left(\frac{2\pi}{2(n + 2)}\right)} \]

### Step 6: Simplify the expression
After simplification, we can see that:
\[ S = 0 \]
because the sine terms will cancel out due to the periodic nature of the sine function.

### Final Result
Thus, the sum \( S = \sin 2\alpha + \sin 3\alpha + \ldots + \sin n\alpha = 0 \).

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