sin α−sin(1200−α)+sin(1200+α)=
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Answer:
sinxsin(120
sinxsin(120 0
sinxsin(120 0 −x)sin(120
sinxsin(120 0 −x)sin(120 0
sinxsin(120 0 −x)sin(120 0 +x)
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx(
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 4
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)=
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 4
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41 sin3x
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41 sin3xFundamental period of sinax is
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41 sin3xFundamental period of sinax is a
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41 sin3xFundamental period of sinax is a2π
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41 sin3xFundamental period of sinax is a2π
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41 sin3xFundamental period of sinax is a2π
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41 sin3xFundamental period of sinax is a2π So, period of
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41 sin3xFundamental period of sinax is a2π So, period of 4
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41 sin3xFundamental period of sinax is a2π So, period of 41
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41 sin3xFundamental period of sinax is a2π So, period of 41
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41 sin3xFundamental period of sinax is a2π So, period of 41 sin3x is
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41 sin3xFundamental period of sinax is a2π So, period of 41 sin3x is 3
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41 sin3xFundamental period of sinax is a2π So, period of 41 sin3x is 32π
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41 sin3xFundamental period of sinax is a2π So, period of 41 sin3x is 32π
sinxsin(120 0 −x)sin(120 0 +x) =sinx[sin 2 120 0 −sin 2 x]=sinx( 43 −sin 2 x)= 41 sin3xFundamental period of sinax is a2π So, period of 41 sin3x is 32π
hope it's helpful
❤️sam❤️