Answer
Ans. Sin4$\theta$=$8cos^3θsinθ-4cosθsinθ$
Work Step by Step
Given:- By De moivre's theorem
$(cosθ+isinθ)^n=(cosnθ+sinnθ) $ ~Eq.1
If n=4
$(cosθ+isinθ)^4$=$(cos^2θ+2isinθcosθ-sin^2θ)(cos^2θ+2isinθcosθ-sin^2θ)$
=$(cos^4θ-6sin^2θcos^2θ+sin^4θ)+i(4sinθcos^3θ-4sin^3θcosθ)$
After comparing of Eq.1 LHS and RHS we get
Sin4$\theta$=$(4sinθcos^3θ-4sin^3θcosθ)$
Put $sin^2θ=1-cos^2θ$ in the above equation
=$8cos^3θsinθ-4cosθsinθ$
Ans.
