If A + B + c = 180 then show that

sinA + sinB + sinC=4cosA/2. cosB/2 . cosC/2

Answer:

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To prove that sin(A) + sin(B) + sin(C) = 4cos(A/2) * cos(B/2) * cos(C/2), we can use the half-angle formulas for cosine. Let's break it down step by step.

1. Start with the given equation: A + B + C = 180.
This equation represents the angle sum of a triangle.

2. Rewrite the equation in terms of angles A/2, B/2, and C/2.
Dividing both sides of the equation by 2, we get:
A/2 + B/2 + C/2 = 90.

3. Apply the half-angle formula for cosine to each angle:
cos^2(A/2) = (1 + cos(A))/2
cos^2(B/2) = (1 + cos(B))/2
cos^2(C/2) = (1 + cos(C))/2

4. Rearrange the equations from step 3 to solve for cos(A), cos(B), and cos(C):
cos(A) = 2cos^2(A/2) - 1
cos(B) = 2cos^2(B/2) - 1
cos(C) = 2cos^2(C/2) - 1

5. Substitute the values of cos(A), cos(B), and cos(C) into the given equation:
sin(A) + sin(B) + sin(C) = 4cos(A/2) * cos(B/2) * cos(C/2)

Plugging in the values from step 4, we have:
sin(A) + sin(B) + sin(C) = 4[(2cos^2(A/2) - 1)/2][(2cos^2(B/2) - 1)/2][(2cos^2(C/2) - 1)/2]

6. Simplify the expression:
sin(A) + sin(B) + sin(C) = 4(cos^2(A/2) - 1/2)(cos^2(B/2) - 1/2)(cos^2(C/2) - 1/2)

Applying the identity cos^2(x) = 1 - sin^2(x), we get:
sin(A) + sin(B) + sin(C) = 4[(1 - sin^2(A/2))/2][(1 - sin^2(B/2))/2][(1 - sin^2(C/2))/2]

Simplifying further:
sin(A) + sin(B) + sin(C) = 4[(1/2)(1 - sin^2(A/2))][(1/2)(1 - sin^2(B/2))][(1/2)(1 - sin^2(C/2))]

Finally, using the identity sin^2(x) = 1 - cos^2(x), we have:
sin(A) + sin(B) + sin(C) = 4(cos(A/2) * cos(B/2) * cos(C/2))

And that concludes the proof that sin(A) + sin(B) + sin(C) = 4cos(A/2) * cos(B/2) * cos(C/2).