Answer to Question #114076 in Algebra for Azwinndini

Question

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) [30]

Answer 1

10.1 Let

$\theta = \frac {\pi} {10} = 18 ^ 0\\ 5 \theta = \frac {\pi} {2}\\ \cos3 \theta = \sin2 \theta \\ \cos54 ^ 0 = \sin36 ^ 0 = \sin (90 ^ 0-54 ^ 0) \\ 4 \cos ^ 3 \theta-3 \cos \theta = 2 \sin \theta \cos \theta \\ 4 \cos ^ 2 \theta-3 = 2 \sin \theta \\ 4 (1- \sin ^ 2 \theta) -3 = 2 \sin \theta \\ 4 \sin ^ 2 \theta + 2 \sin \theta-1 = 0 \\ \sin \theta = \frac {\sqrt5-1} {4} \\ \cos2 \theta = \cos \frac {\pi} {5} = 1-2 \sin ^ 2 \theta = \frac {\sqrt5 + 1} {4} \\ \cos \frac {\pi} {5}= \frac {\sqrt5 + 1} {4}\\ \sin \frac {\pi} {5} = \sqrt {1- \cos ^ 2 \theta} = \sqrt {1 - (\frac {\sqrt5 + 1} {4}) ^ 2} = \frac {\sqrt {10-2 \sqrt5}} {4} \\ \sin \frac {2 \pi} {5 } = 2 \sin \frac {\pi} {5} \cos \frac {\pi} {5} = \frac {\sqrt {10-2 \sqrt5}} {4} \frac {\sqrt5 + 1} {2} = \\ = \frac {\sqrt {10 + 2 \sqrt5}} {4}$

10.2

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

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

$(c)\\ \sin \theta = \frac {z-z ^ {- 1}} {2i} \\ \sin ^ 7 \theta = \frac {(z-z ^ {- 1}) ^ 7} {(2i)^7} = \\ = \frac {1} {(2i)^7} (z ^ 7-7z ^ 6z ^ {- 1} + 21z ^ 5z ^ {- 2} -35z ^ 4z ^ {- 3} + \\ + 35z ^ 3z ^ {- 4} - 21z ^ 2z ^ {- 5} + 7zz ^ {- 6} -z ^ {- 7}) = \\ = \frac {1} {(2i)^7} (2i \sin7 \theta-7 \cdot (2i) \sin5 \theta +\\ + 21 \cdot (2i) \sin3 \theta - 35 (2i) \sin \theta) = \\ =\frac{1}{(2i)^6} (\sin7 \theta-7 \sin5 \theta + 21 \sin3 \theta-35 \ sin\theta)=\\ =-\frac{1}{64} (\sin7 \theta-7 \sin5 \theta + 21 \sin3 \theta-35 \ sin\theta)\\$

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

$(е)\\ \cos3 \theta = \frac {z ^ 3 + z ^ {- 3}} {2} = \frac {(z + z ^ {- 1}) (z ^ 2-zz ^ {- 1} + z ^ {-2})} {2} = \\ = \frac {2 \ cos\theta (2 \cos2 \theta-1)} {2} = \cos \theta (4 \cos ^ 2 \theta-3) = \\ = 4 \cos ^ 3 \theta-3 \cos \theta \\ 4x = \cos3 \theta + 3 \cos \theta \\ 4x = 4 \cos ^ 3 \theta-3 \cos \theta + 3 \cos \theta \\ x = \cos ^ 3 \theta\\ \sin3 \theta = \frac {z ^ 3-z ^ {- 3}} {2i} = \frac {(zz ^ {- 1}) (z ^ 2 + zz ^ {- 1} + z ^ {- 2})} {2i} = \\ = \frac {2i \sin \theta (2 \cos2 \theta + 1)} {2i} = \sin \theta (3-4 \ sin ^ 2 \theta) = \ \ = 3 \sin \theta-4 \sin ^ 3 \theta \\ 4y = 3 \sin \theta- \sin3 \theta \\ 4y = 3 \sin \theta-3 \sin \theta + 4 \sin ^ 3 \theta \\ y = \sin ^ 3 \theta$