Question #c7682

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Answer 1 · 2017-11-30T07:03:01

By Eular's formula we have

#cosz+isinz=e^(iz).......[1]#

#cosz-isinz=e^(-iz).......[2]#

Subtracting [2] from [1] we get

#sinz=(e^(iz)-e^(-iz))/(2i)#

#=>sin^3z=[(e^(iz)-e^(-iz))/(2i)]^3#

#=>sin^3z=-1/(8i)[(e^(iz)-e^(-iz))]^3#

#=>-4sin^3z=1/(2i)[(e^(i3z)-e^(-i3z))-3*e^(iz)*e^(-iz)(e^(iz)-e^(-iz))]#

#=>-4sin^3z=(e^(i3z)-e^(-i3z))/(2i)-(3(e^(iz)-e^(-iz)))/(2i)#

#=>-4sin^3z=sin3z-3sinz#

#=>sin3z=3sinz-4sin^3z#