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What is the Value of sina/2 sinb/2 sinc/2?with proof of the question

What is the Value of sina/2 sinb/2 sinc/2?with proof of the question

Grade:12

1 Answers

Vikas TU
14149 Points
6 years ago
Let f = sin(A/2) . sin(B/2) . sin(C/2)
⇒f =sin(A/2) . sin(B/2) . sin(π/2 − A+B/2)   [as, A+B+C = π]
⇒f = sin(A/2) . sin(B/2) . cos(A+B/2)   
  ∂f∂A = sin(B/2)[cos(A+B/2) . cos(A/2) −sin(A/2) . sin(A+B/2)]
⇒∂f∂A = sin(B/2) . cos(2A+B/2)  
 ∂f∂B = sin(A/2)[cos(A+B/2) . cos(B/2) −sin(B/2) . sin(A+B/2)]
⇒∂f∂B = sin(A/2) . cos(A+2B/2)
   Now, r = ∂2f/∂A2 = − sin(B/2) . sin (2A+B/2)
s = ∂2f/∂B ∂A = ∂/∂A(∂f∂B) = cos(A+2B2)   . cos (A2) . 12 − 12 sin(A2) . sin (A+2B2)s = 12cos(A+B)t = ∂2f/∂B2 = − sin(A2) . sin (A+2B2)
Now, ∂f/∂A = 0   and   ∂f/∂B = 0⇒sin(B2) . cos(2A+B2) = 0 and sin(A2) . cos(A+2B2)⇒cos(2A+B2) = 0 and  cos(A+2B2) = 0     [as,sin(A2) ≠0 and sin(B2)≠0]
⇒2A + B = π and A + 2B = π
Solving above 2 equations, we getA = B = π/3
AT A = B = π/3 
now, r = −sin(π6) . sin [2π/3+π/3/2] = −1/2
s = 12cos(π3+π3) = −1/4
t = −sin(π6) . sin [2π/3+π/3/2] = −12
Now, rt−s2 = (−12)(−12) − (−14)2 = 316 > 0
Also r = −12 
So, f is maximum at A = B = π/3
Maximum value of f = f(π/3, π/3) = sin(A/2) . sin(B/2) . cos(A+B/2) = sin(π/6) . sin(π/6) . cos(π/3+π/3/2) = 12×12×12 = 1/8

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