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prove that: sin2(n + 1)A - sin2nA = sin(2n + 1)A.sinA
prove that:
sin2(n + 1)A - sin2nA = sin(2n + 1)A.sinA
sin2(n+1)A - sin2nA = sin(2n+1)A.sinA we have formula sin2a - sin2b =sin(a-b)sin(a+b) therefore sin2(n+1)A - sin2nA=sin[(n+1)A+nA].sin[(n+1)A-nA] =sin(2n+1)A.sinA =RHS hence proved
sin2(n+1)A - sin2nA = sin(2n+1)A.sinA
we have formula sin2a - sin2b =sin(a-b)sin(a+b)
therefore
sin2(n+1)A - sin2nA=sin[(n+1)A+nA].sin[(n+1)A-nA]
=sin(2n+1)A.sinA =RHS
hence proved
sin^2(n + 1)A - sin^2nA = sin(2n + 1)A.sinASolution:- L.H.S:-sin^2(n + 1)A - sin^2nA = Sin[(n+1)A+nA].Sin[(n+1)A-nA] ["a^2-b^2=(a+b)(a-b)"] =Sin[nA+A+nA].Sin[nA+A-nA] =Sin[A(n+1+n)].SinA =Sin(2n+1)A.SinA=R.H.S Thank You!!!!!!!!!!!!!!!! Wish You Happy learning of trignometry..............
sin^2(n + 1)A - sin^2nA = sin(2n + 1)A.sinASolution:- L.H.S:-sin^2(n + 1)A - sin^2nA = Sin[(n+1)A+nA].Sin[(n+1)A-nA] ["a^2-b^2=(a+b)(a-b)"] =Sin[nA+A+nA].Sin[nA+A-nA] =Sin[A(n+1+n)].SinA =Sin(2n+1)A.SinA=R.H.S
Thank You!!!!!!!!!!!!!!!!
Wish You Happy learning of trignometry..............
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