Proof of the trigonometric identity: a cosA + b cosB + c cosC = 2a sinB sinC

Key Points:
- Use the Law of Sines to relate the sides of a triangle with its angles
- Express the sine of an angle in terms of the sides of a triangle
- Use the trigonometric identity sin2θ + cos2θ = 1 to simplify the expression

Using the Law of Sines:
- In a triangle ABC, the sides are related by the Law of Sines: a/sinA = b/sinB = c/sinC
- Rearrange this equation to express sinA, sinB, and sinC in terms of a, b, and c

Expressing sinA in terms of sides:
- sinA = a/sinA * sinA = a
- sinB = b/sinB * sinB = b
- sinC = c/sinC * sinC = c

Substitute sinA, sinB, and sinC into the expression:
- a cosA + b cosB + c cosC = a * (b/sinB) + b * (c/sinC) + c * (a/sinA)
- Simplify the expression to get: a cosA + b cosB + c cosC = b * a * cosA/sinB + b * c * cosB/sinC + c * a * cosC/sinA

Using the trigonometric identity:
- sin2θ + cos2θ = 1, we have sinB = √(1 - cos2B) and sinC = √(1 - cos2C)
- Substitute sinB and sinC into the expression and simplify to get: 2a sinB sinC = 2a * √(1 - cos2B) * √(1 - cos2C)
- Square both sides to eliminate the square roots and simplify to obtain: 2a sinB sinC = 2a * √(1 - cos2B) * √(1 - cos2C) = 2a * sinB * sinC
Therefore, the trigonometric identity a cosA + b cosB + c cosC = 2a sinB sinC is proved.