sin³x+sin³(120°+x)+sin³(240°+x)=?

প্রশ্ন: sin³x+sin³(120°+x)+sin³(240°+x)=?

উত্তর: -¾sin3x

সমাধান:

আমরা জানি, sin³θ = (3sinθ - sin3θ)/4

সুতরাং,

sin³x = (3sinx - sin3x)/4

sin³(120°+x) = [3sin(120°+x) - sin(3(120°+x))]/4 = [3sin(120°+x) - sin(360°+3x)]/4 = [3sin(120°+x) - sin3x]/4

sin³(240°+x) = [3sin(240°+x) - sin(3(240°+x))]/4 = [3sin(240°+x) - sin(720°+3x)]/4 = [3sin(240°+x) - sin3x]/4

এখন, sin³x+sin³(120°+x)+sin³(240°+x) = (3sinx - sin3x)/4 + [3sin(120°+x) - sin3x]/4 + [3sin(240°+x) - sin3x]/4

= [3sinx + 3sin(120°+x) + 3sin(240°+x) - 3sin3x]/4

= [3{sinx + sin(120°+x) + sin(240°+x)} - 3sin3x]/4

আমরা জানি, sinA + sinB = 2sin((A+B)/2)cos((A-B)/2)

সুতরাং, sin(120°+x) + sin(240°+x) = 2sin((120°+x+240°+x)/2)cos((120°+x-240°-x)/2) = 2sin(180°+x)cos(-60°) = 2(-sinx)(1/2) = -sinx

অতএব, sinx + sin(120°+x) + sin(240°+x) = sinx - sinx = 0

তাহলে, [3{sinx + sin(120°+x) + sin(240°+x)} - 3sin3x]/4 = [3(0) - 3sin3x]/4 = -3sin3x/4 = -¾sin3x

সুতরাং, sin³x+sin³(120°+x)+sin³(240°+x) = -¾sin3x প্রমাণিত। 🎉