how sin2a+sin2b+sin2c is equal to 4 sin a sin b sin c
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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
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