How do you prove: sin3θ = 3sinθ−4sin3θ?

The left-hand side as sin3θ=sin(θ+2θ)

now expand the right side of this equation using Addition formula

sin(A±B)= sinAcosB ± cosAsinB sin(θ+2θ)=sinθcos2θ+cosθsin2θ…….(1)

cos2θ=cos2θ−sin2θ=2cos2θ−1=1−2sin2θ

The right hand side is expressed only in terms of sin θ ‘ s

so we use cos 2 θ = 1 − 2 sin2θ … … . . (2 )

sin 2 θ = 2 sin θ cos θ …….(3)

Replace cos 2 θ and sin 2 θ by the expansions (2) and (3) into (1)

sin ( θ + 2θ ) = sin θ ( 1 − 2 sin 2 θ ) + cos θ ( 2 sin θ cos θ ) and expanding brackets gives.

sin ( θ + 2θ ) = sin θ − 2 sin3θ + 2sinθcos2θ … . (4 )

cos 2θ + sin 2θ = 1

⇒ cos2θ = 1 − sin 2 θ

Replace cos2θ = 1 − sin2θ into (4)

sin (θ + 2θ) = sin θ − 2 sin3θ + 2sinθ ( 1 − sin2θ) and

expanding 2nd bracket gives. sin (θ + 2 sin θ ) = sin θ − 2 sin3θ + 2sinθ − 2sin3θ

Terms are collected

sin 3θ = 3sinθ − 4 sin3θ = R.H.S hence proven

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