10- Using De Moivre's theorem to prove that:
\[
\sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta, \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta
\]
Answer
The simplified expressions are: \(\boxed{\sin(3\theta) = 3\sin(\theta) - 4\sin(\theta)^3}\) and \(\boxed{\cos(3\theta) = 4\cos(\theta)^3 - 3\cos(\theta)}\)
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 \(\boxed{\cos(3\theta) = 4\cos(\theta)^3 - 3\cos(\theta)}\)
