Arun
Last Activity: 6 Years ago
i) Let x = y = z = k -------- (1)
ii) Substituting in the given other two equations from (1) and reducing, we get:
sin(A) + sin(B) + sin(C) = 2d²/k ------ (2)
and sin(2A) + sin(2B) + sin(2C) = d²/k -------- (3)
iii) Dividing (3) by (2):
[sin(2A) + sin(2B) + sin(2C)]/[sin(A) + sin(B) + sin(C)] = 1/2 ---------- (4)
iv) You may be aware from trigonometric identities for the given condition A + B + C = π, that
sin(2A) + sin(2B) + sin(2C) = 4*sin(A)*sin(B)*sin(C) and
sin(A) + sin(B) + sin(C) = 4*cos(A/2)*cos(B/2)*cos(C/2)
Substituting these in (4) above,
4*sin(A)*sin(B)*sin(C)/ 4*cos(A/2)*cos(B/2)*cos(C/2) = 1/2
==> sin(A)*sin(B)*sin(C)/cos(A/2)*cos(B/2)*c... = 1/2
==> [2sin(A/2)cos(A/2)*2sin(B/2)cos(B/2)*2si... = 1/2
{Application of multiple angle identities}
On cancellation of the corresponding terms and simplification we get,
sin(A/2)*sin(B/2)*sin(C/2) = 1/16
