Demonstrate that in the triangle ABC sinA + sinB + sinC < 3(sqrt3)/2.
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To prove sinA+sinB+sinC = 3(sqrt3)/2.
We know that sinA+sinB = sin(a+b)/2 * cos(A-B)/2.
Sinc = sin (180-(A+B)) = sin(A+B) = sin(A+B) = 2in(A+B)/2 cos(A+B)/2
Therefore sinA+sinB +sinC = 2sin(A+B)/2 * (cosA-B)/2 + 2sin(A+B)/ *cos (A*B)/2 = 2 sin (A+B)/2 * {cos(A-B)/2 +cos (A+B)/2}
= 2 sin(A+B)/2 {2Cos A/2 * cosB/2}
= 4sin(A+B)/2 }{cos(90-(B+C)/2) cos(90-(A+B)/2)
= 4 sin (A+B)/2 sin (B+C)/2 sin(C+A)/2.
So if x+y+z =1 and xyz is maximum when x=y=z =1/3.
So A+B = B+C = A+C. A =B= C = (A+B+C) /3, when sinA+sin B+ sin C is maximum = 4 sin {(180/3)}^3 = 4(sin60)^3 = (sqrt3/2)^3 = 4*3sqrt3/2^3 = 3sqrt3/2.
Therefore sinA+sinB+sinC < (3sqrt3)/2
We'll choose to set the function f(x) = sin x, on the bracket [0,pi](we don't have to forget that we are working in a triangle, where the sum of the angles is 180 degrees, meaning pi, if we're measuring the angles in radians.)
f'(x) = cos x, f"(x) = - sin x < 0, so the function is concave and we we'll apply the Jensen's inequality, which says that:
f[(A + B + C)/3] > [f(A) + f(B) + f(C)]/3
Working in a triangle, A + B + C = 180 and (A + B + C)/3 = 180/3 = 60
f(60) = sin 60 = (sqrt 3)/2
[f(A) + f(B) + f(C)]/3 = (sin A + sin B + sin C)/3
Putting the results again in Jensen's inequality:
(sin A + sin B + sin C)/3 < (sqrt 3)/2
sin A + sin B + sin C < 3(sqrt 3)/2 q.e.d.
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