10.1
Let θ = π 10 = 1 8 0 \theta=\frac{\pi}{10}=18^0 θ = 10 π = 1 8 0 5 θ = π 2 5\theta=\frac{\pi}{2} 5 θ = 2 π
cos 3 θ = sin 2 θ cos 5 4 0 = sin 3 6 0 = sin ( 9 0 0 − 5 4 0 ) 4 cos 3 θ − 3 cos θ = 2 sin θ cos θ 4 cos 2 θ − 3 = 2 sin θ 4 ( 1 − sin 2 θ ) − 3 = 2 sin θ 4 sin 2 θ + 2 sin θ − 1 = 0 sin θ = 5 − 1 4 cos 2 θ = cos π 5 = 1 − 2 sin 2 θ = 5 + 1 4 sin π 5 = 1 − cos 2 θ = 1 − ( 5 + 1 4 ) 2 = 10 − 2 5 4 sin 2 π 5 = 2 sin π 5 cos π 5 = 10 − 2 5 4 5 + 1 2 = = 10 + 2 5 4 \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} cos 3 θ = sin 2 θ cos 5 4 0 = sin 3 6 0 = sin ( 9 0 0 − 5 4 0 ) 4 cos 3 θ − 3 cos θ = 2 sin θ cos θ 4 cos 2 θ − 3 = 2 sin θ 4 ( 1 − sin 2 θ ) − 3 = 2 sin θ 4 sin 2 θ + 2 sin θ − 1 = 0 sin θ = 4 5 − 1 cos 2 θ = cos 5 π = 1 − 2 sin 2 θ = 4 5 + 1 sin 5 π = 1 − cos 2 θ = 1 − ( 4 5 + 1 ) 2 = 4 10 − 2 5 sin 5 2 π = 2 sin 5 π cos 5 π = 4 10 − 2 5 2 5 + 1 = = 4 10 + 2 5
10.2
z = cos θ + i sin θ z n = cos n θ + i sin n θ z − n = 1 cos n θ + i sin n θ = cos n θ − i sin n θ n ∈ N 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 z = cos θ + i sin θ z n = cos n θ + i sin n θ z − n = c o s n θ + i s i n n θ 1 = cos n θ − i sin n θ n ∈ N
(a)
z n + z − n = 2 cos n θ z n − z − n = 2 i sin n θ z^n+z^{-n}=2\cos n\theta\\
z^n-z^{-n}=2i\sin n\theta z n + z − n = 2 cos n θ z n − z − n = 2 i sin n θ
(b)
z = cos θ + i sin θ ( z + 1 ) n = ( cos θ + i sin θ + 1 ) n = = ( 2 cos 2 θ 2 + i 2 sin θ 2 cos θ 2 ) n = = 2 n cos n θ 2 ( cos θ 2 + i sin θ 2 ) n = = 2 n cos n θ 2 ( cos n θ 2 + i sin n θ 2 ) = = 2 n cos n θ 2 z n 2 2 n cos n θ 2 z n 2 = 2 n cos n θ 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 = cos θ + i sin θ ( z + 1 ) n = ( cos θ + i sin θ + 1 ) n = = ( 2 cos 2 2 θ + i 2 sin 2 θ cos 2 θ ) n = = 2 n cos n 2 θ ( cos 2 θ + i sin 2 θ ) n = = 2 n cos n 2 θ ( cos 2 n θ + i sin 2 n θ ) = = 2 n cos n 2 θ z 2 n 2 n cos n 2 θ z 2 n = 2 n cos n θ
( z − 1 ) n = ( cos θ + i sin θ − 1 ) n = = ( 2 sin 2 θ 2 + i 2 sin θ 2 cos θ 2 ) n = = 2 n sin n θ 2 ( sin θ 2 + i cos θ 2 ) n = = 2 n sin n θ 2 ( − i ( cos θ 2 + i sin θ 2 ) ) n = = ( 2 i ) n sin n θ 2 ( cos n θ 2 + i sin n θ 2 ) = = ( 2 i ) n sin n θ 2 z n 2 ( 2 i ) n sin n θ 2 z n 2 = ( 2 i ) n sin n θ (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\\ ( z − 1 ) n = ( cos θ + i sin θ − 1 ) n = = ( 2 sin 2 2 θ + i 2 sin 2 θ cos 2 θ ) n = = 2 n sin n 2 θ ( sin 2 θ + i cos 2 θ ) n = = 2 n sin n 2 θ ( − i ( cos 2 θ + i sin 2 θ ) ) n = = ( 2 i ) n sin n 2 θ ( cos 2 n θ + i sin 2 n θ ) = = ( 2 i ) n sin n 2 θ z 2 n ( 2 i ) n sin n 2 θ z 2 n = ( 2 i ) n sin n θ
(c)
sin θ = z − z − 1 2 i sin 7 θ = ( z − z − 1 ) 7 2 i = = 1 2 i ( z 7 − 7 z 6 z − 1 + 21 z 5 z − 2 − 35 z 4 z − 3 + + 35 z 3 z − 4 − 21 z 2 z − 5 + 7 z z − 6 − z − 7 ) = = 1 2 i ( 2 i sin 7 θ − 7 ⋅ ( 2 i ) sin 5 θ + 21 ⋅ ( 2 i ) sin 3 θ − − 35 ( 2 i ) sin θ ) = = s i n 7 θ − 7 sin 5 θ + 21 sin 3 θ − 35 sin θ \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 sin θ = 2 i z − z − 1 sin 7 θ = 2 i ( z − z − 1 ) 7 = = 2 i 1 ( z 7 − 7 z 6 z − 1 + 21 z 5 z − 2 − 35 z 4 z − 3 + + 35 z 3 z − 4 − 21 z 2 z − 5 + 7 z z − 6 − z − 7 ) = = 2 i 1 ( 2 i sin 7 θ − 7 ⋅ ( 2 i ) sin 5 θ + 21 ⋅ ( 2 i ) sin 3 θ − − 35 ( 2 i ) sin θ ) = = s in 7 θ − 7 sin 5 θ + 21 sin 3 θ − 35 sin θ
(d)
cos 3 θ = z 3 + z − 3 2 = ( z + z − 1 ) ( z 2 − z z − 1 + z − 2 ) 2 = = 2 cos θ ( 2 cos 2 θ − 1 ) 2 = cos θ ( 2 cos 2 θ − 1 ) sin 4 θ = z 4 − z − 4 2 i = ( z 2 + z − 2 ) ( z 2 − z − 2 ) 2 i = 2 cos 2 θ 2 i sin 2 θ 2 i = = 2 cos 2 θ sin 2 θ \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 cos 3 θ = 2 z 3 + z − 3 = 2 ( z + z − 1 ) ( z 2 − z z − 1 + z − 2 ) = = 2 2 c o s θ ( 2 c o s 2 θ − 1 ) = cos θ ( 2 cos 2 θ − 1 ) sin 4 θ = 2 i z 4 − z − 4 = 2 i ( z 2 + z − 2 ) ( z 2 − z − 2 ) = 2 i 2 c o s 2 θ 2 i s i n 2 θ = = 2 cos 2 θ sin 2 θ
(e)
cos 3 θ = z 3 + z − 3 2 = ( z + z − 1 ) ( z 2 − z z − 1 + z − 2 ) 2 = = 2 cos θ ( 2 cos 2 θ − 1 ) 2 = cos θ ( 4 cos 2 θ − 3 ) = = 4 cos 3 θ − 3 cos θ 4 x = cos 3 θ + 3 cos θ 4 x = 4 cos 3 θ − 3 cos θ + 3 cos θ x = cos 3 θ \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 cos 3 θ = 2 z 3 + z − 3 = 2 ( z + z − 1 ) ( z 2 − z z − 1 + z − 2 ) = = 2 2 c o s θ ( 2 c o s 2 θ − 1 ) = cos θ ( 4 cos 2 θ − 3 ) = = 4 cos 3 θ − 3 cos θ 4 x = cos 3 θ + 3 cos θ 4 x = 4 cos 3 θ − 3 cos θ + 3 cos θ x = cos 3 θ
sin 3 θ = z 3 − z − 3 2 i = ( z − z − 1 ) ( z 2 + z z − 1 + z − 2 ) 2 i = = 2 i sin θ ( 2 cos 2 θ + 1 ) 2 i = sin θ ( 3 − 4 sin 2 θ ) = = 3 sin θ − 4 sin 3 θ 4 y = 3 sin θ − sin 3 θ 4 y = 3 sin θ − 3 sin θ + 4 sin 3 θ y = sin 3 θ \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 sin 3 θ = 2 i z 3 − z − 3 = 2 i ( z − z − 1 ) ( z 2 + z z − 1 + z − 2 ) = = 2 i 2 i s i n θ ( 2 c o s 2 θ + 1 ) = sin θ ( 3 − 4 sin 2 θ ) = = 3 sin θ − 4 sin 3 θ 4 y = 3 sin θ − sin 3 θ 4 y = 3 sin θ − 3 sin θ + 4 sin 3 θ y = sin 3 θ
#339914 on Dec 2023
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