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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

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

Grade:12

4 Answers

sanjay sanjeev patro
43 Points
8 years ago

THE PROBLEM OR FORMULA IS RIGHT ,SO

sin3x=3sinx-4sin3x

JAYANT ARORA
37 Points
8 years ago

 The derivation of the formula :- sin(3x)=3sinx - 4sin3x is absolutely correct . . . 

Rohan Das
36 Points
8 years ago

this is the derivation of sin3x.i thnk all the steps are correct.

vishal ojha
35 Points
8 years ago

sin 3x =3sin x -4sin3x

cos3x = 4 cos3x- 3 cosx

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