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Class 10 > Learn Class IX-X Tuition > IF A+B+C = Pie Then prove that Sin2A + Sin2B + Sin2C...
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IF A+B+C = Pie Then prove that Sin2A + Sin2B + Sin2C = 4SinASinBSinC U.K.AKSHAY

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sin2A + sin 2B +sin2C = 2 sin(A+B)cos(A-B) + 2sinC cosC =2sinC cos(A-B)+2sinC cosC =2sinC (cos(A-b) + cos C) =2sin C(cos(A-B) - cos(A+B) ) = 2sinC . 2sin A sin B =4 sinA sin B sin C
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LHS = sin 2A + sin 2B + sin 2C = sin (pie -- 2A) + sin (pie --2B) + sin (pie --2C) = 2sin(A+B)cos(A-B) + 2sinCcosC = 2sin(pie - C)cos(A-B) + 2sinCcosC = 2sinCcos(A - B) + 2sinCcosC = 2sinC (cos (A - B)+ cos C) = 2sinC (2cos (A...
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LHS = sin 2A + sin 2B + sin 2C = sin (pie – 2A) + sin (pie –2B) + sin (pie –2C) = 2sin(A+B)cos(A-B) + 2sinCcosC = 2sin(pie - C)cos(A-B) + 2sinCcosC = 2sinCcos(A - B) + 2sinCcosC = 2sinC (cos (A - B)+ cos C) = 2sinC (2cos (A - B + C)/2 cos (A - B - C)/2) = 2sinC (2cos (pie - 2B)/2 cos(2A - pie)/2 = 4sinAsinBsinC = RHS Hence Proved read less
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Math Educator for Std.11th ,12th , Engineering Entrance and Degree Level with 11+ Years Experience

sin2A + sin2B + sin2C = 2 sin (2A + 2B)/2 . cos (2A - 2B)/2 + sin2C = 2sin(A + B).cos(A - B) + 2 sinC.cosC = 2sin(A + B).cos(A - B) + 2 sin (Pie - (A + B)) cos (Pie - (A + B)) = 2 sin(A + B) (cos (A - B) - cos (A + B)) = 2 sin(A + B).2sinA.sinB = 4 sinA.sinB.sinC
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Sin 2 is a common so, take those common factor out sin2(A+B+C), we know that A+B+C=pie apply on that so, sin2pie is equal to in value 0.1096
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LHS 2sinAcosA+2sinBcosB+2sinCcosC....(1) A= pi-(B+C), B= pi-(A+C), C= pi-(B+A). So, cosA= cos = -cos = - ......(2) cosB= cos = -cos = - cosC= cos =.-cos = - . Putting these values in (1) and combining common terms -2 + 6sinAsinBsinC. putting value for cosBcosC from (2) in...
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LHS 2sinAcosA+2sinBcosB+2sinCcosC....(1) A= pi-(B+C), B= pi-(A+C), C= pi-(B+A). So, cosA= cos[pi-(B+C)] = -cos[B+C] = - [cosBcosC-sinBsinC] ......(2) cosB= cos[pi-(A+C)] = -cos[A+C] = - [cosAcosC-sinAsinC] cosC= cos[pi-(B+A)] =.-cos[B+A] = - [cosBcosA-sinBsinA]. Putting these values in (1) and combining common terms -2[sinAcosBcosC+sinBcosCcosA+sinCcosBcosA] + 6sinAsinBsinC. putting value for cosBcosC from (2) in first term and taking cosA common from second and 3rd term = -2[sinA [sinBsinc-cosA] + cosA [sin (B+C)] ] + 6sinAsinBsinC. As A= pi-(B+C) sinA=sin(B+C) we get after cancellation = -2 [sinAsinBsinC] + 6sinAsinBsinC. = 4sinAsinBsinC.= RHS Hence proved read less
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Experienced mathematics faculty

sin2A + sin 2B +sin2C = 2 sin(A+B)cos(A-B) + 2sinC cosC =2sinC cos(A-B)+2sinC cosC =2sinC (cos(A-b) + cos C) =2sin C(cos(A-B) - cos(A+B) ) = 2sinC . 2sin A sin B =4 sinA sin B sin C
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sin2A + sin2B + sin2C = 2 sin (2A + 2B)/2 . cos (2A - 2B)/2 + sin2C = 2sin(A + B).cos(A - B) + 2 sinC.cosC = 2sin(A + B).cos(A - B) + 2 sin (Pie - (A + B)) cos (Pie - (A + B)) = 2 sin(A + B) (cos (A - B) - cos (A + B)) = 2 sin(A + B).2sinA.sinB = 4 sinA.sinB.sinC
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LHS = sin 2A + sin 2B + sin 2C = sin (pie -- 2A) + sin (pie --2B) + sin (pie --2C) = 2sin(A+B)cos(A-B) + 2sinCcosC = 2sin(pie - C)cos(A-B) + 2sinCcosC = 2sinCcos(A - B) + 2sinCcosC = 2sinC (cos (A - B)+ cos C) = 2sinC (2cos (A - B + C)/2 cos (A - B - C)/2) = 2sinC (2cos (pie - 2B)/2 cos(2A...
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LHS = sin 2A + sin 2B + sin 2C = sin (pie – 2A) + sin (pie –2B) + sin (pie –2C) = 2sin(A+B)cos(A-B) + 2sinCcosC = 2sin(pie - C)cos(A-B) + 2sinCcosC = 2sinCcos(A - B) + 2sinCcosC = 2sinC (cos (A - B)+ cos C) = 2sinC (2cos (A - B + C)/2 cos (A - B - C)/2) = 2sinC (2cos (pie - 2B)/2 cos(2A - pie)/2 = 4sinAsinBsinC read less
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Excellence Award Winner ( 7 times )

2sinAcosA+2sinBcosB+2sinCcosC....(1) A= pi-(B+C), B= pi-(A+C), C= pi-(B+A). So, cosA= cos = -cos = - ......(2) cosB= cos = -cos = - cosC= cos =.-cos = - . Putting these values in (1) and combining common terms -2 + 6sinAsinBsinC. putting value for cosBcosC from (2) in first...
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2sinAcosA+2sinBcosB+2sinCcosC....(1) A= pi-(B+C), B= pi-(A+C), C= pi-(B+A). So, cosA= cos[pi-(B+C)] = -cos[B+C] = - [cosBcosC-sinBsinC] ......(2) cosB= cos[pi-(A+C)] = -cos[A+C] = - [cosAcosC-sinAsinC] cosC= cos[pi-(B+A)] =.-cos[B+A] = - [cosBcosA-sinBsinA]. Putting these values in (1) and combining common terms -2[sinAcosBcosC+sinBcosCcosA+sinCcosBcosA] + 6sinAsinBsinC. putting value for cosBcosC from (2) in first term and taking cosA common from second and 3rd term = -2[sinA [sinBsinc-cosA] + cosA [sin (B+C)] ] + 6sinAsinBsinC. As A= pi-(B+C) sinA=sin(B+C) we get after cancellation = -2 [sinAsinBsinC] + 6sinAsinBsinC. = 4sinAsinBsinC.= RHS - read less
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Left Hand Side 2sinAcosA+2sinBcosB+2sinCcosC....(1) A= pi-(B+C), B= pi-(A+C), C= pi-(B+A). So, cosA= cos = -cos = - ......(2) cosB= cos = -cos = - cosC= cos =.-cos = - . Putting these values in (1) and combining common terms -2 + 6sinAsinBsinC. putting value for cosBcosC...
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Left Hand Side 2sinAcosA+2sinBcosB+2sinCcosC....(1) A= pi-(B+C), B= pi-(A+C), C= pi-(B+A). So, cosA= cos[pi-(B+C)] = -cos[B+C] = - [cosBcosC-sinBsinC] ......(2) cosB= cos[pi-(A+C)] = -cos[A+C] = - [cosAcosC-sinAsinC] cosC= cos[pi-(B+A)] =.-cos[B+A] = - [cosBcosA-sinBsinA]. Putting these values in (1) and combining common terms -2[sinAcosBcosC+sinBcosCcosA+sinCcosBcosA] + 6sinAsinBsinC. putting value for cosBcosC from (2) in first term and taking cosA common from second and 3rd term = -2[sinA [sinBsinc-cosA] + cosA [sin (B+C)] ] + 6sinAsinBsinC. As A= pi-(B+C) sinA=sin(B+C) we get after cancellation = -2 [sinAsinBsinC] + 6sinAsinBsinC. = 4sinAsinBsinC. = Right Hand Side read less
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