Question

Let z = cosθ + isinθ.
Then z^n = cos(nθ) + isin(nθ) for all n ∈ N(by de Moivre) and z^−n = cos(nθ)−isin(nθ).

(a) Show that 2cos(nθ) = z^n + z^−n and 2isin(nθ) = z^n −z^−n

(b) Show that 2^ncos^n θ = (z + 1/z)^n and (2i)^n sin^n θ = (z− 1 z/n)

(c) Use (b) to express sin^7 θ in terms of multiple angles

(d) Express cos^3 θsin^4 θ in terms of multiple angles

(e) Eliminate θ from the equations 4x = cos(3θ) + 3cosθ; 4y = 3sinθ−sin(3θ).

Answer 1

$\bold{Solution} \\ \bold {a)}\\ z^{n} + z^{-n} = cos(nθ)+ i sin(nθ) + cos(nθ) - i sin(nθ) = 2cos(nθ) \\ z^{n} - z^{-n} = cos(nθ)+ i sin(nθ) - cos(nθ) + i sin(nθ) =2isin(nθ) \\ \bold{b)} \\ From \ a)\ follow \ that \ when \ n=1 \\ z^{1} + z^{−1}= 2cos(θ)\\ ( z^1 + z^{−1})^n =(2cos(θ) )^n = 2^n cos^n(θ)\\ z^1 - z^{−1}= 2i sin(θ)\\ ( z^1 - z^{−1})^n =(2i sin(θ) )^n = (2i)^n sin^n(θ) \\ \bold{c) }\\ From \ b) \ we\ have \ that\\ (2i sinθ)^7 = (z-z^{-1})^7 = C_{0 }^7 z^7 (-(z^{-1}))^0 + C_{1}^7 z^6 (-(z^{-1}))^1 + C_{2}^7 z^5 (-(z^{-1}))^2+C_{3}^7 z^4 (-(z^{-1}))^3+ C_{4}^7 z^3 (-(z^{-1}))^4 + C_{5}^7 z^2 (-(z^{-1}))^5 +C_{6}^7 z^1 (-(z^{-1}))^6 +C_{7}^7 z^0 (-(z^{-1}))^7 =\\ = 1z^7 - 7z^6 z^{-1} + 21z^5 z^{-2} - 35z^4 z^{-3} +35z^3 z^{-4} - 21z^2 z^{-5} + 7z^1 z^{-6}-1z^0 z^{-7} =\\ = z^7 – 7 z^5+ 21 z^3 – 35z + 35 z^{-1} – 21z^{-3} + 7z^{-5} - z^{-7}=\\ = ( z^7 - z^{-7}) – 7( z^5 - z^{-5}) +21( z^3-z^{-3} ) – 35(z - z^{-1} ) =\\ =2isin⁡7θ – 7*2isin⁡5θ + 21* 2isin⁡3θ – 35*2isin⁡θ =\\ = 2i (sin⁡7θ – 7sin⁡5θ + 21sin⁡3θ – 35sin⁡θ)\\ (2i)^7 (sinθ)^7 =(2i sinθ)^7\\ sin^7 θ=\frac{sin⁡7θ – 7 sin⁡5θ + 21 sin⁡3θ – 35 sin⁡θ}{-64}\\ \bold{d)}\\ 2^3 cos^3 θ = (z^1+ z^{-1})^3 = 1z^3 + 3z^2 z^{-1} + 3 z^1 z^{-2} + 1z^0 z^{-3} =\\ = z^3+3z+3z^{-1}+ z^{-3} =\\ = (z^3+ z^{-3} ) +3(z^1+ z^{-1}) =\\ = 2cos⁡3θ + 3 *2cos⁡θ =\\ = 2cos⁡3θ + 6cos⁡θ\\ cos^3 θ = \frac{1}{4} cos⁡3θ + \frac{3}{4} cos⁡θ\\ (2i)^3 sin^4 θ = (z^1-z^{-1})^4 = 1z^4 + 4z^3 (-z^{-1})^1 + 6 z^2 (-z^{-1})^2 + 4 z^1 (-z^{-1})^3 + 1 (-z^{-1})^4 =\\ = z^4-4z^2 +6 – 4 z^{-2} + z^{-4} =\\ = (z^4+ z^{-4})-4(z^2+ z^{-2})+ 6 =\\ = 2cos⁡4θ – 4*2cos2⁡θ+6\\ sin^4 θ= \frac{1}{8} cos⁡4θ- \frac{1}{2} cos2⁡θ+\frac{3}{8}\\ cos^3 θ sin^4 θ= (\frac{1}{4} cos⁡(3θ) + \frac{3}{4} cos⁡(θ))(\frac{1}{8} cos⁡(4θ)- \frac{1}{2} cos(2⁡θ)+\frac{3}{8}) =\frac{cos(-\theta)+cos(7\theta)}{64} - \frac{cos(\theta)+cos(5\theta)}{16} + \frac{3}{32}cos(3\theta) + \frac{3(cos(-3\theta)+cos(5\theta))}{64} - \frac{3(cos(-\theta)+cos(3\theta))}{16} +\frac{9}{32}cos(\theta)=3cos(\theta)-3cos(3\theta)-cos(5\theta)+cos(7\theta)\\ \bold{e)}\\ From \ a)\ we\ have\ that\\ 4x = cos⁡3θ + 3 cos⁡θ\\ 4y = 3sin⁡θ - sin⁡3θ\\ cos⁡3θ = \frac{z^3+ z^{-3}}{2} = \frac{(z^1+ z^{-1})(z^2-z z^{-1}+ z^{-2}))}{2} = \\ =\frac{2 cos⁡θ (2 cos2⁡θ-1))}{2} = cos⁡θ(4cos^2 θ – 3) =\\ = 4cos^3 θ – 3cos⁡θ\\ 4x = 4cos^3 θ\\ x = cos^3 θ\\ cos⁡θ=\sqrt[3]{x}\\ \theta=arccos(\sqrt[3]{x})\\ sin⁡3θ = \frac{z^3- z^{-3}}{2} = \frac{(z^1- z^{-1})(z^2+z z^{-1}+ z^{-2})}{2i} = \\ =\frac{2i sin⁡θ (2 cos2⁡θ+1)}{2i}= sin⁡θ(3 - 4sin^2 θ ) = 3sin⁡θ - 4sin^3 θ\\ 4y = 4sin^3 θ\\ y = sin^3 θ\\ sin⁡θ=\sqrt[3]{y}\\ \theta=arcsin(\sqrt[3]{y})\\$