Mumbai University > First Year Engineering > Sem 1 > Applied Maths 1

Marks : 6 M

Year : May 2014

c$os6θ+isin6θ \;=\; (cosθ+isinθ)^6 \; \; \ldots By \; De \; Moivre's \; Theorem \\ cosnθ+isinnθ \;=\; (cosθ+isinθ)^n \\ \; \\ \; \\ = cos^6θ + 6cos^5θ(isinθ)+15cos^4θ(isinθ)^2+20cos^3θ(isinθ)^3 \\ + 15cos^2θ(isinθ)^4 + 6cosθ(isinθ)^5 + (isinθ)^6 \\ \; \\ \; \\ = cos^6θ + 6icos^5θsinθ-15cos^4θsin^2θ-20icos^3θsin^3θ \\ + 15cos^2θsin^4θ + 6icosθsin^5θ - sin^6θ \\ \; \\ \; \\ = ( cos^6θ -15cos^4θsin^2θ+ 15cos^2θsin^4θ - sin^6θ) \\ + i( 6cos^5θsinθ-20cos^3θsin^3θ +6cosθsin^5θ ) \\ \; \\ \; \\ $

Comparing real and imaginary parts, we get

$ sin6\theta \;=\; 6cos^5θsinθ-20cos^3θsin^3θ +6cosθsin^5θ \\ \; \\ \; \\ \therefore \dfrac{sin6\theta}{sin\theta} \;=\; 6cos^5θ-20cos^3θsin^2θ +6cosθsin^4θ \\ \; \\ \; \\ = 6cos^5θ-20cos^3θ(1-cos^2θ) +6cosθ(1-cos^2θ)^2 \\ \; \\ \ ;\\ = 6cos^5θ-20cos^3θ+20cos^5θ +6cosθ(1 -2cos^2θ +cos^4θ) \\ \; \\ \ ;\\ = 6cos^5θ-20cos^3θ+20cos^5θ +6cosθ -12cos^3θ +6cos^5θ \\ \; \\ \; \\ = 32cos^5θ-32cos^3θ+6cosθ \\ \; \\ \; \\ = 2cos\theta(16cos^4θ-16cos^2θ+3) \\ \; \\ \; \\ \therefore \dfrac{sin6\theta}{2sin\theta cos\theta} \;=\; 16cos^4θ-16cos^2θ+3 \\ \; \\ \; \\ \; \\ \dfrac{sin6\theta}{sin2\theta } \;=\; 16cos^4θ-16cos^2θ+3 \; \; \; Hence \; Proved. $