Let
θ = π / 5 , = > 5 θ = π . 3 θ = π − 2 θ . \theta=\pi/5, =>5\theta=\pi.\\
3\theta=\pi-2\theta. \\ θ = π /5 , => 5 θ = π . 3 θ = π − 2 θ . Therefore,
sin ( 3 θ ) = sin ( π − 2 θ ) = sin ( 2 θ ) \text{sin}(3\theta)=\text{sin}(\pi-2\theta)=\text{sin}(2\theta) sin ( 3 θ ) = sin ( π − 2 θ ) = sin ( 2 θ ) if θ < 90. \theta<90. θ < 90.
Next,
sin 3 θ = sin 2 θ , 3 sin θ − 4 sin 3 θ = 2 sin θ cos θ ∣ ∣ ⋅ 1 sin θ , 3 − 4 sin 2 θ = 2 cos θ , 3 − 4 ( 1 − cos 2 θ ) = 2 cos θ , 4 cos 2 θ − 2 cos θ − 1 = 0. \text{sin}3\theta=\text{sin}2\theta,\\
3\text{sin}\theta-4\text{sin}^3\theta=2\text{sin}\theta\text{cos}\theta\space||\cdot\frac{1}{\text{sin}\theta},\\
3-4\text{sin}^2\theta=2\text{cos}\theta,\\
3-4(1-\text{cos}^2\theta)=2\text{cos}\theta,\\
4\text{cos}^2\theta-2\text{cos}\theta-1=0. sin 3 θ = sin 2 θ , 3 sin θ − 4 sin 3 θ = 2 sin θ cos θ ∣∣ ⋅ sin θ 1 , 3 − 4 sin 2 θ = 2 cos θ , 3 − 4 ( 1 − cos 2 θ ) = 2 cos θ , 4 cos 2 θ − 2 cos θ − 1 = 0. We notice that this is a quadratic equation where cos theta is like a variable:
cos θ = 1 ± 5 4 , \text{cos}\theta=\frac{1\pm\sqrt5}{4}, cos θ = 4 1 ± 5 , we take only positive root because theta cannot give negative cosine,
cos θ = cos π 5 = 1 + 5 4 , sin θ = sin π 5 = 1 − cos 2 θ = 10 − 2 5 4 . sin 2 π 5 = sin ( 2 θ ) = 2 sin θ cos θ = = 2 ⋅ 10 − 2 5 4 ⋅ 1 + 5 4 = 10 + 2 5 4 . \text{cos}\theta=\text{cos}\frac{\pi}{5}=\frac{1+\sqrt5}{4},\\
\text{sin}\theta=\text{sin}\frac{\pi}{5}=\sqrt{1-\text{cos}^2\theta}=\frac{\sqrt{10-2\sqrt5}}{4}.\\
\text{sin}\frac{2\pi}{5}=\text{sin}(2\theta)=2\text{sin}\theta\text{cos}\theta=\\
=2\cdot\frac{\sqrt{10-2\sqrt5}}{4}\cdot\frac{1+\sqrt5}{4}
=\frac{\sqrt{{10+2\sqrt5}}}{4}. cos θ = cos 5 π = 4 1 + 5 , sin θ = sin 5 π = 1 − cos 2 θ = 4 10 − 2 5 . sin 5 2 π = sin ( 2 θ ) = 2 sin θ cos θ = = 2 ⋅ 4 10 − 2 5 ⋅ 4 1 + 5 = 4 10 + 2 5 .
(a)
z n + z − n = ( cos θ + i sin θ ) n + ( cos θ + i sin θ ) − n , z^n+z^{-n}=(\text{cos}\theta+i\text{sin}\theta)^n+(\text{cos}\theta+i\text{sin}\theta)^{-n},\\ z n + z − n = ( cos θ + i sin θ ) n + ( cos θ + i sin θ ) − n , according to de Moivre theorem this is
cos ( n θ ) + i sin ( n θ ) + cos ( − n θ ) + i sin ( − n θ ) = = cos ( n θ ) + i sin ( n θ ) + cos ( − n θ ) − i sin ( n θ ) = = 2 cos ( n θ ) . \text{cos}(n\theta)+i\text{sin}(n\theta)+\text{cos}(-n\theta)+i\text{sin}(-n\theta)=\\
=\text{cos}(n\theta)+i\text{sin}(n\theta)+\text{cos}(-n\theta)-i\text{sin}(n\theta)=\\
=2\text{cos}(n\theta). cos ( n θ ) + i sin ( n θ ) + cos ( − n θ ) + i sin ( − n θ ) = = cos ( n θ ) + i sin ( n θ ) + cos ( − n θ ) − i sin ( n θ ) = = 2 cos ( n θ ) . 
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