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If the line x=y=z intersect the line sinA x+sinB y+sinC z=2d^2,sin2A x+sin2B y+sin2C z=d^2 then sinA/2sinB/2sinC/2 is equal to where (A+B+C=pi)

If the line x=y=z intersect the line sinA x+sinB y+sinC z=2d^2,sin2A x+sin2B y+sin2C z=d^2 then sinA/2sinB/2sinC/2 is equal to where (A+B+C=pi)
 

Grade:11

1 Answers

Arun
25763 Points
2 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

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