How do you express sin 3theta in terms of trigonometric functions of theta?

We need to find the formula for sin 3Θ


We know that

sin (x + y) = sin x cos y + cos x sin y

cos (Θ + y) = cos x cos y – sin x sin y

We can write sin Θ

sin 3Θ = sin (2Θ + Θ)

= sin 2Θ cos Θ + cos 2Θsin Θ

=> sin (Θ + Θ)* cos Θ + cos (Θ +Θ)*sin Θ

=> [sin Θ* cos Θ + cos Θ*sin Θ]* cos Θ + [cos Θ* cos Θ – sin Θ*sin Θ]*sin Θ

=> 2*sin Θ * (cos Θ)2 + (cos Θ)2 * sin Θ – (sin Θ)3

=> 3sin Θ (cos Θ)2 – (sin Θ)3

Now (cos Θ)2 + (sin Θ)2= 1

or we can also express it as (cos Θ)2 = 1 – (sin Θ)2

=> 3* sin Θ*(1 – (sin Θ)2) – (sin Θ)3

=> 3* sin Θ – 3(sin Θ)2) – (sin Θ)3

=> 3*sin Θ – 4*(sin Θ)3

Therefore we get

sin 3Θ = 3sin Θ – 4*(sin Θ)3

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