Let a, b,c are respectively the sines and p, q, r are respectively the consines of `alpha, alpha+(2pi)/(3) and alpha+(4pi)/(3)`, then :
Q. The value of `(a+b+c)` is :

Options

A. 0
B. `(3)/(4)`
C. 1
D. none of these

Answer

To solve the problem, we need to find the value of \( a + b + c \), where \( a = \sin(\alpha) \), \( b = \sin\left(\alpha + \frac{2\pi}{3}\right) \), and \( c = \sin\left(\alpha + \frac{4\pi}{3}\right) \).

### Step 1: Define the values
Let:
- \( a = \sin(\alpha) \)
- \( b = \sin\left(\alpha + \frac{2\pi}{3}\right) \)
- \( c = \sin\left(\alpha + \frac{4\pi}{3}\right) \)

### Step 2: Write the expression for \( a + b + c \)
We need to calculate:
\[ a + b + c = \sin(\alpha) + \sin\left(\alpha + \frac{2\pi}{3}\right) + \sin\left(\alpha + \frac{4\pi}{3}\right) \]

### Step 3: Use the sine addition formula
To simplify \( b \) and \( c \), we can use the sine addition formula:
\[ \sin(x + y) = \sin x \cos y + \cos x \sin y \]

Calculating \( b \) and \( c \):
- For \( b \):
\[ b = \sin\left(\alpha + \frac{2\pi}{3}\right) = \sin(\alpha)\cos\left(\frac{2\pi}{3}\right) + \cos(\alpha)\sin\left(\frac{2\pi}{3}\right) \]
Using \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \) and \( \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \):
\[ b = \sin(\alpha)\left(-\frac{1}{2}\right) + \cos(\alpha)\left(\frac{\sqrt{3}}{2}\right) = -\frac{1}{2}\sin(\alpha) + \frac{\sqrt{3}}{2}\cos(\alpha) \]

- For \( c \):
\[ c = \sin\left(\alpha + \frac{4\pi}{3}\right) = \sin(\alpha)\cos\left(\frac{4\pi}{3}\right) + \cos(\alpha)\sin\left(\frac{4\pi}{3}\right) \]
Using \( \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} \) and \( \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2} \):
\[ c = \sin(\alpha)\left(-\frac{1}{2}\right) + \cos(\alpha)\left(-\frac{\sqrt{3}}{2}\right) = -\frac{1}{2}\sin(\alpha) - \frac{\sqrt{3}}{2}\cos(\alpha) \]

### Step 4: Combine \( a + b + c \)
Now substituting \( b \) and \( c \) into \( a + b + c \):
\[ a + b + c = \sin(\alpha) + \left(-\frac{1}{2}\sin(\alpha) + \frac{\sqrt{3}}{2}\cos(\alpha)\right) + \left(-\frac{1}{2}\sin(\alpha) - \frac{\sqrt{3}}{2}\cos(\alpha)\right) \]

### Step 5: Simplify the expression
Combining the terms:
\[ a + b + c = \sin(\alpha) - \frac{1}{2}\sin(\alpha) - \frac{1}{2}\sin(\alpha) + \frac{\sqrt{3}}{2}\cos(\alpha) - \frac{\sqrt{3}}{2}\cos(\alpha) \]
The \( \sin(\alpha) \) terms cancel out:
\[ = \sin(\alpha) - \sin(\alpha) = 0 \]

### Conclusion
Thus, the value of \( a + b + c \) is:
\[ \boxed{0} \]