If `a = cos theta + isin theta`, `b = cosphi + isinphi`, `c = cos psi + i sin psi` and `a/b+b/c+c/a=2` then `sin(theta-phi)+sin(phi-psi)+sin(psi-theta)` equals

Answer

To solve the given problem step by step, we start with the expressions for \( a \), \( b \), and \( c \):

1. **Define the complex numbers**:
\[ a = \cos \theta + i \sin \theta = e^{i \theta} \]
\[ b = \cos \phi + i \sin \phi = e^{i \phi} \]
\[ c = \cos \psi + i \sin \psi = e^{i \psi} \]

2. **Set up the equation**:
We are given that:
\[ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 2 \]
Substituting the expressions for \( a \), \( b \), and \( c \):
\[ \frac{e^{i \theta}}{e^{i \phi}} + \frac{e^{i \phi}}{e^{i \psi}} + \frac{e^{i \psi}}{e^{i \theta}} = 2 \]
This simplifies to:
\[ e^{i(\theta - \phi)} + e^{i(\phi - \psi)} + e^{i(\psi - \theta)} = 2 \]

3. **Use the properties of complex exponentials**:
We know that \( e^{ix} = \cos x + i \sin x \). Therefore, the equation can be rewritten as:
\[ \left(\cos(\theta - \phi) + i \sin(\theta - \phi)\right) + \left(\cos(\phi - \psi) + i \sin(\phi - \psi)\right) + \left(\cos(\psi - \theta) + i \sin(\psi - \theta)\right) = 2 \]

4. **Separate real and imaginary parts**:
The real part must equal 2 and the imaginary part must equal 0:
\[ \cos(\theta - \phi) + \cos(\phi - \psi) + \cos(\psi - \theta) = 2 \]
\[ \sin(\theta - \phi) + \sin(\phi - \psi) + \sin(\psi - \theta) = 0 \]

5. **Analyze the imaginary part**:
From the equation \( \sin(\theta - \phi) + \sin(\phi - \psi) + \sin(\psi - \theta) = 0 \), we can deduce that the sum of the sines is zero.

6. **Conclusion**:
Therefore, we find that:
\[ \sin(\theta - \phi) + \sin(\phi - \psi) + \sin(\psi - \theta) = 0 \]

Thus, the final answer is:
\[ \sin(\theta - \phi) + \sin(\phi - \psi) + \sin(\psi - \theta) = 0 \]