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

Marks : 6 M

Year : May 2015

We know that

$(cos⁡7θ+i sin⁡7θ )=〖(cos⁡θ+i sin⁡θ)^7 \ldots By \; De Moivre' s \; theorem \\ \; \\ \; \\ \therefore (cos⁡7θ+i sin⁡7θ ) \;=\; \\ cos^7 θ+7cos^6 θi sin⁡θ+21cos^5 θi^2 sin^2 θ+35cos^4 θi^3 sin^3 θ+35cos^3 θi^4 sin^4 θ \\ \; \\ \; \\ \therefore (cos⁡7θ+i sin⁡7θ ) \;=\; cos^7 θ+i(7cos^6 θ sin⁡θ)-21cos^5 θsin^2 θ-i(35cos^4 θsin^3 θ) \\ +35cos^3 θsin^4 θ +i(21cos^2 θsin^5 θ)-7 cos⁡θ sin^6 θ-isin^7 θ \\ \; \\ \; \\ $

Separating Real and Imaginary Parts,

$ \therefore (cos⁡7θ+i sin⁡7θ )=(cos^7 θ-21cos^5 θsin^2 θ+35cos^3 θsin^4 θ-7 cos⁡θ sin^6 θ) \\ +i(7cos^6 θ sin⁡θ-35cos^4 θsin^3 θ+21cos^2 θsin^5 θ-sin^7 θ) \\ \; \\ \; \\ \therefore sin⁡7θ=7cos^6 θ sin⁡θ-35cos^4 θsin^3 θ+21cos^2 θsin^5 θ-sin^7 θ \\ =7(cos^2 θ)^3 sin⁡θ-35(cos^2 θ)^2 sin^3 θ+21(1-sin^2 θ)sin^5 θ-sin^7 θ \\ \; \\ \; \\ \therefore sin⁡7θ=7 (1-sin^2 θ)^3 sin⁡θ-35(1-sin^2 θ)^2 sin^2 θ sin⁡θ +21(1-sin^2 θ) sin^4 θ sin⁡θ-sin^6 θ sin⁡θ \\ \; \\ \; \\ \therefore \dfrac{sin7\theta}{sin\theta} =7(1-sin^2 θ)^3-35(1-sin^2 θ)^2 sin^2 θ+ 21(1-sin^2 θ) sin^4 θ-sin^6 θ \\ \; \\ \; \\ \therefore \dfrac{sin7\theta}{sin\theta} =7(1-3sin^2 θ+3sin^4 θ-sin^6 θ) \\ -35(1-2sin^2 θ+sin^4 θ) sin^2 θ+21sin^4 θ-21sin^6 θ-sin^6 θ \\ \; \\ \; \\ \therefore \dfrac{sin7\theta}{sin\theta} =7-21sin^2 θ+21sin^4 θ-7sin^6 θ-35sin^2 θ+ 70sin^4 θ-35sin^6 θ+21sin^4 θ-21sin^6 θ-sin^6 θ \\ \; \\ \; \\ \; \\ \therefore \dfrac{sin7\theta}{sin\theta} =7-56sin^2 θ+112sin^4 θ-64sin^6 θ $