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sin(2x+x)=sin2x.cosx+cos2x.sinx
=2sinxcosx.cosx+(-2sin^2x)sinx
=2sinxcos^2+sinx-2sin^3x
=sinx(2cos^2x+1)-2sin^3x
=sinx(2-2sin^2x+1)-2sin^3x
=2sinx-2sin^3x+sinx-2sin^3x
= 3sinx-4sin^3x
THE PROBLEM OR FORMULA IS RIGHT ,SO
sin3x=3sinx-4sin3x
The derivation of the formula :- sin(3x)=3sinx - 4sin3x is absolutely correct . . .
this is the derivation of sin3x.i thnk all the steps are correct.
sin 3x =3sin x -4sin3x
cos3x = 4 cos3x- 3 cosx
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