Question

prove that :- sin 3tita = 3 sin tita - 4 sin³tita

Answer 1

Explanation:

Express the left hand side as

sin3θ=sin(θ+2θ)

now expand the right side of this equation using Addition formula

∣∣ ∣ ∣ ∣ ∣ ∣ ∣∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aasin(A±B)=sinAcosB±cosAsinBaa∣∣−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

⇒sin(θ+2θ)=sinθcos2θ+cosθsin2θ.......(A)

∣∣ ∣ ∣ ∣ ∣ ∣ ∣∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aacos2θ=cos2θ−sin2θ=2cos2θ−1=1−2sin2θaa∣∣−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

The right hand side is expressed only in terms of sinθ's

so we use cos2θ=1−2sin2θ........(1)

∣∣ ∣∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aasin2θ=2sinθcosθaa∣∣−−−−−−−−−−−−−−−−−−−−−........(2)

Replace cos2θ and sin2θ by the expansions (1) and (2) 
into (A)

sin(θ+2θ)=sinθ(1−2sin2θ)+cosθ(2sinθcosθ)

and expanding brackets gives.

sin(θ+2θ)=sinθ−2sin3θ+2sinθcos2θ....(B)

∣∣ ∣ ∣ ∣ ∣ ∣ ∣∣¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯aacos2θ+sin2θ=1⇒cos2θ=1−sin2θaa∣∣−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

Replace cos2θ=1−sin2θ into (B)

⇒sin(θ+2θ)=sinθ−2sin3θ+2sinθ(1−sin2θ)

and expanding 2nd bracket gives.

sin(θ+2sinθ)=sinθ−2sin3θ+2sinθ−2sin3θ

Finally, collecting like terms.

sin3θ=3sinθ−4sin3θ=R.H.S hence proven