Answer to Question #113414 in Linear Algebra for Mzwandile

Question

QUESTION 10

10.1 Use 9.2 to evaluate sin π/5 , sin 2π/5 and cos π/5 .

10.2 Let z = cos θ + i sin θ.

Then zn = cos(nθ) + i sin(nθ) for all n ∈ N (by de Moivre) and z−n = cos(nθ) − i sin(nθ).

(a) Show that 2 cos(nθ) = zn + z−n and 2i sin(nθ) = zn − z−n.
(2)

(b) Show that 2n cosn θ = (z + 1 )n and (2i)n sinn θ = (z − 1 )n.
(2)



(c) Use (b) to express sin7 θ in terms of multiple angles.
(6)

(d) Express cos3 θ sin4 θ in terms of multiple angles.
(6)

(e) Eliminate θ from the equations 4x = cos(3θ) + 3 cos θ; 4y = 3 sin θ − sin(3θ).
(5)

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Answer 1

10.1

Let $\theta=\frac{\pi}{10}=18^0$ then $5\theta=\frac{\pi}{2}$

$\cos3\theta=\sin2\theta\\ \cos54^0=\sin36^0=\sin(90^0-54^0)\\ 4\cos^3\theta-3\cos\theta=2\sin\theta\cos\theta\\ 4\cos^2\theta-3=2\sin\theta\\ 4(1-\sin^2\theta)-3=2\sin\theta\\ 4\sin^2\theta+2\sin\theta-1=0\\ \sin\theta=\frac{\sqrt5-1}{4}\\ \cos2\theta=\cos\frac{\pi}{5}=1-2\sin^2\theta=\frac{\sqrt5+1}{4}\\ \sin\frac{\pi}{5}=\sqrt{1-\cos^2\theta}=\sqrt{1-(\frac{\sqrt5+1}{4})^2}=\frac{\sqrt{10-2\sqrt5}}{4}\\ \sin\frac{2\pi}{5}=2\sin\frac{\pi}{5}\cos\frac{\pi}{5}=\frac{\sqrt{10-2\sqrt5}}{4}\frac{\sqrt5+1}{2}=\\ =\frac{\sqrt{10+2\sqrt5}}{4}$

10.2

$z=\cos\theta+i\sin\theta\\ z^n=\cos n\theta+i\sin n\theta\\ z^{-n}=\frac{1}{\cos n\theta+i\sin n\theta}=\cos n\theta-i\sin n\theta\\ n\in N$

(a)

$z^n+z^{-n}=2\cos n\theta\\ z^n-z^{-n}=2i\sin n\theta$

(b)

$z=\cos\theta+i\sin\theta\\ (z+1)^n=(\cos\theta+i\sin\theta+1)^n=\\ =(2\cos^2\frac{\theta}{2}+i2\sin\frac{\theta}{2}\cos\frac{\theta}{2})^n=\\ =2^n\cos^n\frac{\theta}{2}(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2})^n=\\ =2^n\cos^n\frac{\theta}{2}(\cos\frac{n\theta}{2}+i\sin\frac{n\theta}{2})=\\ =2^n\cos^n\frac{\theta}{2}z^{\frac{n}{2}}\\ 2^n\cos^n\frac{\theta}{2}z^{\frac{n}{2}}=2^n\cos^ n\theta\\$

$(z-1)^n=(\cos\theta+i\sin\theta-1)^n=\\ =(2\sin^2\frac{\theta}{2}+i2\sin\frac{\theta}{2}\cos\frac{\theta}{2})^n=\\ =2^n\sin^n\frac{\theta}{2}(\sin\frac{\theta}{2}+i\cos\frac{\theta}{2})^n=\\ =2^n\sin^n\frac{\theta}{2}(-i(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}))^n=\\ =(2i)^n\sin^n\frac{\theta}{2}(\cos\frac{n\theta}{2}+i\sin\frac{n\theta}{2})=\\ =(2i)^n\sin^n\frac{\theta}{2}z^{\frac{n}{2}}\\ (2i)^n\sin^n\frac{\theta}{2}z^{\frac{n}{2}}=(2i)^n\sin^ n\theta\\$

(c)

$\sin\theta=\frac{z-z^{-1}}{2i}\\ \sin^7\theta=\frac{(z-z^{-1})^7}{2i}=\\ =\frac{1}{2i}(z^7-7z^6z^{-1}+21z^5z^{-2}-35z^4z^{-3}+\\+35z^3z^{-4} -21z^2z^{-5}+7zz^{-6}-z^{-7})=\\ =\frac{1}{2i}(2i\sin7\theta-7\cdot(2i)\sin5\theta+21\cdot(2i)\sin3\theta-\\ -35(2i)\sin\theta)=\\ =sin7\theta-7\sin5\theta+21\sin3\theta-35\sin\theta$

(d)

$\cos3\theta=\frac{z^3+z^{-3}}{2}=\frac{(z+z^{-1})(z^2-zz^{-1}+z^{-2})}{2}=\\ =\frac{2\cos\theta(2\cos2\theta-1)}{2}=\cos\theta(2\cos2\theta-1)\\ \sin4\theta=\frac{z^4-z^{-4}}{2i}=\frac{(z^2+z^{-2})(z^2-z^{-2})}{2i}=\frac{2\cos2\theta2i\sin2\theta}{2i}=\\ =2\cos2\theta\sin2\theta$

(e)

$\cos3\theta=\frac{z^3+z^{-3}}{2}=\frac{(z+z^{-1})(z^2-zz^{-1}+z^{-2})}{2}=\\ =\frac{2\cos\theta(2\cos2\theta-1)}{2}=\cos\theta(4\cos^2\theta-3)=\\ =4\cos^3\theta-3\cos\theta\\ 4x=\cos3\theta+3\cos\theta\\ 4x=4\cos^3\theta-3\cos\theta+3\cos\theta\\ x=\cos^3\theta$

$\sin3\theta=\frac{z^3-z^{-3}}{2i}=\frac{(z-z^{-1})(z^2+zz^{-1}+z^{-2})}{2i}=\\ =\frac{2i\sin\theta(2\cos2\theta+1)}{2i}=\sin\theta(3-4\sin^2\theta)=\\ =3\sin\theta-4\sin^3\theta\\ 4y=3\sin\theta-\sin3\theta\\ 4y=3\sin\theta-3\sin\theta+4\sin^3\theta\\ y=\sin^3\theta$