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( cos  + i sin) 3 = ( cos  + i sin) ( cos  + i sin) ( cos  + i sin)

To Prove: Cos 3  =c os 3  - 3 sin 2  cos  and Sin3  = sin  - 4 sin 3 . Proof:. ( cos  + i sin) 3 = ( cos  + i sin) ( cos  + i sin) ( cos  + i sin). = ( cos 2  + i sin cos  + i sin  cos  + i 2 sin 2 )( cos  + i sin).

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( cos  + i sin) 3 = ( cos  + i sin) ( cos  + i sin) ( cos  + i sin)

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  1. To Prove: Cos 3 =cos3 - 3sin2cosand Sin3 = sin - 4 sin3 Proof: (cos + i sin)3= (cos + i sin) (cos + i sin) (cos + i sin) = ( cos2 + i sin cos + i sin  cos  + i2 sin2)(cos  + i sin) = ( cos2  + 2i sin cos  - sin2 )(cos  + i sin ) = cos3 + 2i sincos2 - sin2cos + i cos2 sin + 2i2 sin2cos - i sin3 =cos3 + 2i sincos2 - sin2cos + i cos2 sin - 2 sin2cos - i sin3 = cos3 - sin2cos - 2 sin2cos+i(2 sincos2 + cos2 sin - sin3) = cos3 - 3sin2cos +i(2 sincos2 + cos2 sin - sin3) But (cos + i sin)3 = cos 3 + i sin 3 (De Moivre’s Theorem) Hence cos 3 + i sin 3 = cos3 - 3 sin2 cos + i(3 sin cos2 - sin3) Real = Real and Imaginary=Imaginary cos 3 =cos3 - 3sin2cos sin 3 =3 sincos2 - sin3 = 3 sin (1 - sin 2 ) - sin3 sin 3 = sin - 4sin3 Q.E.D. (c) Project Maths Development Team 2011

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