In a triangle ABC, if a, b, c are in A.P. and `(b)/(c) sin 2C + (c)/(b) sin 2B + (b)/(a) sin 2A + (a)/(b) sin 2B = 2`, then find the value of sin B

Explanation

To solve the problem step-by-step, we will use the information given in the question along with some trigonometric identities and properties of triangles.

### Step 1: Understanding the Given Information
We are given that in triangle ABC, the sides \( a, b, c \) are in Arithmetic Progression (A.P.). This means:
\[ 2b = a + c \]

### Step 2: Using the Sine Rule
According to the sine rule in a triangle:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
where \( R \) is the circumradius of the triangle. We can express the sides in terms of the angles:
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]

### Step 3: Substitute the Values of a, b, and c
Using the relationship \( 2b = a + c \):
\[ 2(2R \sin B) = 2R \sin A + 2R \sin C \]
Dividing by \( 2R \) (assuming \( R \neq 0 \)):
\[ 2 \sin B = \sin A + \sin C \]

### Step 4: Analyzing the Given Equation
The equation given in the problem is:
\[ \frac{b}{c} \sin 2C + \frac{c}{b} \sin 2B + \frac{b}{a} \sin 2A + \frac{a}{b} \sin 2B = 2 \]
Substituting \( a, b, c \) in terms of \( R \) and the sines of the angles:
\[ \frac{2R \sin B}{2R \sin C} \sin 2C + \frac{2R \sin C}{2R \sin B} \sin 2B + \frac{2R \sin B}{2R \sin A} \sin 2A + \frac{2R \sin A}{2R \sin B} \sin 2B = 2 \]
This simplifies to:
\[ \frac{\sin B}{\sin C} \sin 2C + \frac{\sin C}{\sin B} \sin 2B + \frac{\sin B}{\sin A} \sin 2A + \frac{\sin A}{\sin B} \sin 2B = 2 \]

### Step 5: Simplifying the Equation
We can rewrite the equation:
\[ \sin B \sin 2C + \sin C \sin 2B + \sin B \sin 2A + \sin A \sin 2B = 2 \sin A \sin B \sin C \]
Factoring out common terms and using the identity \( \sin 2x = 2 \sin x \cos x \):
\[ \sin B \cdot 2 \sin C \cos C + \sin C \cdot 2 \sin B \cos B + \sin B \cdot 2 \sin A \cos A + \sin A \cdot 2 \sin B \cos B = 2 \]

### Step 6: Using the Sine Addition Formula
The terms can be rearranged and simplified using the sine addition formulas:
\[ \sin B \cos C + \sin C \cos B = \sin(B + C) \]
And similar for other terms.

### Step 7: Finalizing the Equation
After simplification, we find:
\[ \sin B + \sin C = 1 \]
Using the earlier result \( 2 \sin B = \sin A + \sin C \), we can substitute to find:
\[ 2 \sin B = 1 \implies \sin B = \frac{1}{2} \]

### Conclusion
Thus, the value of \( \sin B \) is:
\[ \boxed{\frac{1}{2}} \]