Problem

10- Using De Moivre's theorem to prove that: \[ \sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta, \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta \]

Solution

Step 1 :Use De Moivre's theorem to find the expressions for sin(3θ) and cos(3θ) by setting z = cos(θ) + i*sin(θ) and n = 3

Step 2 :Expand (cos(θ) + i*sin(θ))^3 and extract the real and imaginary parts to find the expressions for cos(3θ) and sin(3θ)

Step 3 :Rewrite the expressions for sin(3θ) and cos(3θ) in terms of sin(θ) and cos(θ) only

Step 4 :The simplified expressions are: \(\boxed{\sin(3\theta) = 3\sin(\theta) - 4\sin(\theta)^3}\) and \(\boxed{\cos(3\theta) = 4\cos(\theta)^3 - 3\cos(\theta)}\)

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