Use De Moivre's theorem to express cos 5θ and sin 5θ in terms of sin θ and cos θ.

From the de Moivre's we have,

(cosθ+isinθ) ^n=cos (nθ) + isin (nθ)

Therefore,

R ((cosθ+isinθ) ^5) = cos (5θ) I ((cosθ+isinθ) ^5) = sin (5θ)

Simplifying,

cos^5 (θ) - 10 (sin^2 (θ)) (cos^3 (θ)) + 5 (sin^4 (θ)) (cosθ) = cos (5θ)