In triangle ABC a²sin2B - b²sin2A then find the value of it
In triangle ABC a²sin2B - b²sin2A then find the value of it
Gopal Reddy , 7 Years ago
Grade 10
Trigonometry> In triangle ABC a²sin2B - b²sin2A then fi...
1 Answers
Arun
a²sin2B + b²sin2A = c²cotC/2
4R².sin²A.sin2B + 4R².sin²B.sin2A = 4R².sin²C.cotC/2
sin²A.sin2B + sin²B.sin2A = sin²C.cotC/2
(1 - cos2A)/2.sin2B + (1 - cos2B)/2.sin2A = sin²C.cotC/2
sin2B - cos2A.sin2B + sin2A - cos2B.sin2A = 2sin²C.cotC/2
sin2A + sin2B - sin(2A + 2B) = 2sin²C.cotC/2 ( vì sin2A.cos2B + cos2A.sin2B = sin (2A +2B)
sin2A + sin2B + sin2C = 2sin²C.cotC/2
2sin(A + B)cos(A - B) + 2sinC.cosC = 2sin²C.cotC/2
2sinC.[ cos(A - B) - cos(A + B) ] = 2sin²C.cotC/2
2sinC.-2.sinA.sin(-B) = 2sin²C.cotC/2
4sinA.sinB.sinC = 2sin²C.cotC/2
2sinA.sinB = sinC.cotC/2
2sinA.sinB = 2sinC/2.cosC/2.cosC/2 /sinC/2
sinA.sinB = cos²C/2
2sinA.sinB = 1 + cosC
2sinA.sinB = 1 - cos(A + B)
2sinA.sinB = 1 - cosA.cosB + sinA.sinB
cosA.cosB + sinA.sinB = 1
cos(A - B) = 1
A = B A - B =0
4R².sin²A.sin2B + 4R².sin²B.sin2A = 4R².sin²C.cotC/2
sin²A.sin2B + sin²B.sin2A = sin²C.cotC/2
(1 - cos2A)/2.sin2B + (1 - cos2B)/2.sin2A = sin²C.cotC/2
sin2B - cos2A.sin2B + sin2A - cos2B.sin2A = 2sin²C.cotC/2
sin2A + sin2B - sin(2A + 2B) = 2sin²C.cotC/2 ( vì sin2A.cos2B + cos2A.sin2B = sin (2A +2B)
sin2A + sin2B + sin2C = 2sin²C.cotC/2
2sin(A + B)cos(A - B) + 2sinC.cosC = 2sin²C.cotC/2
2sinC.[ cos(A - B) - cos(A + B) ] = 2sin²C.cotC/2
2sinC.-2.sinA.sin(-B) = 2sin²C.cotC/2
4sinA.sinB.sinC = 2sin²C.cotC/2
2sinA.sinB = sinC.cotC/2
2sinA.sinB = 2sinC/2.cosC/2.cosC/2 /sinC/2
sinA.sinB = cos²C/2
2sinA.sinB = 1 + cosC
2sinA.sinB = 1 - cos(A + B)
2sinA.sinB = 1 - cosA.cosB + sinA.sinB
cosA.cosB + sinA.sinB = 1
cos(A - B) = 1
A = B A - B =0
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