三角函数
先回顾下Cosine 和 Sine 函数定义。先给定一个单位圆(半径是1)的点 C C C，x轴正方向逆时针到那个点的角度为 θ \theta θ
- cos θ \cos{\theta} cosθ 是那个点 x x x轴的坐标
-
sin
θ
\sin{\theta}
sinθ 是那个点
y
y
y轴的坐标
如下图所示：
点 C C C的坐标是 ( x , y ) = ( cos θ , sin θ ) (x,y) = (\cos{\theta}, \sin{\theta}) (x,y)=(cosθ,sinθ)
根据勾股定理， x 2 + y 2 = 1 x^2+y^2=1 x2+y2=1, 所以
cos 2 θ + sin 2 θ = 1 \cos^2{\theta} + \sin^2{\theta} = 1 cos2θ+sin2θ=1
其他公式：
cos
(
θ
1
+
θ
2
)
=
cos
θ
1
cos
θ
2
−
sin
θ
1
sin
θ
2
sin
(
θ
1
+
θ
2
)
=
sin
θ
1
cos
θ
2
+
cos
θ
1
sin
θ
2
\begin{aligned} \cos(\theta_1 + \theta_2) &= \cos{\theta_1}\cos{\theta_2} - \sin{\theta_1} \sin{\theta_2} \\ \sin(\theta_1 + \theta_2) &= \sin{\theta_1} \cos{\theta_2} + \cos{\theta_1} \sin{\theta_2} \end{aligned}
cos(θ1+θ2)sin(θ1+θ2)=cosθ1cosθ2−sinθ1sinθ2=sinθ1cosθ2+cosθ1sinθ2
复数
简单回顾一下复数。一个复数
z
z
z是由两部分组成，实部 和 虚部。
z
=
a
+
b
i
a
,
b
∈
R
,
i
2
=
−
1
z = a + bi \space\space\space\space\space a, b\in \R, \space\space i^2=-1
z=a+bi a,b∈R, i2=−1
其中，
a
,
b
a,b
a,b都是实数，
a
a
a是实部，
b
b
b是虚部。
i
i
i是虚数单位 （
i
2
=
−
1
,
i
=
−
1
i^2 = -1, i = \sqrt{-1}
i2=−1,i=−1
）。
实部是横向轴，虚部是纵向轴，这样一个复数就可以放到2维坐标平面上了。
例如
3
+
4
i
3+4i
3+4i
将实部和虚部从复数提取出来,
ℜ
(
z
)
\Re(z)
ℜ(z) 实部，
ℑ
(
z
)
\Im(z)
ℑ(z) 虚部
ℜ
(
z
)
=
1
2
(
z
+
z
ˉ
)
ℑ
(
z
)
=
1
2
i
(
z
−
z
ˉ
)
\begin{aligned} \Re(z) &= \frac{1}{2}(z + \bar{z}) \\ \Im(z) &= \frac{1}{2i}(z - \bar{z}) \end{aligned}
ℜ(z)ℑ(z)=21(z+zˉ)=2i1(z−zˉ)
我们可以将第一小节中三角函数放到复平面中来
cos
θ
+
i
sin
θ
\cos{\theta}+ i\sin{\theta}
cosθ+isinθ
横轴是
cos
θ
\cos\theta
cosθ，纵轴是坐标是
sin
θ
\sin\theta
sinθ
复数一些运算
两个复数加减
z
1
=
a
1
+
b
1
i
z
2
=
a
2
+
b
2
i
z_1 = a_1 + b_1 i \\ z_2 = a_2 + b_2 i
z1=a1+b1iz2=a2+b2i
z
1
±
z
2
=
(
a
1
±
a
2
)
+
(
b
1
±
b
2
)
i
z_1\pm z_2 = (a_1\pm a_2) + (b_1 \pm b_2) i
z1±z2=(a1±a2)+(b1±b2)i
乘以一个标量 α \alpha α
α ( a + b i ) = α a + α b i \alpha(a+bi) = \alpha a + \alpha b i α(a+bi)=αa+αbi
复数的乘积
z
1
=
a
1
+
b
1
i
z
2
=
a
2
+
b
2
i
z_1 = a_1 + b_1 i \\ z_2 = a_2 + b_2 i
z1=a1+b1iz2=a2+b2i
他们的乘积是
z
1
z
2
=
(
a
1
+
b
1
i
)
(
a
2
+
b
2
i
)
=
a
1
a
2
+
a
1
b
2
i
+
b
1
a
2
i
+
b
1
b
2
i
2
=
(
a
1
a
2
−
b
1
b
2
)
+
(
a
1
b
2
+
b
1
a
2
)
i
\begin{aligned} z_1 z_2 &= (a_1 + b_1 i)(a_2 +b_2 i) \\ &= a_1a_2 + a_1 b_2 i + b_1 a_2 i + b_1 b_2 i^2 \\ &= (a_1 a_2 - b_1 b_2) + (a_1b_2 + b_1 a_2) i \end{aligned}
z1z2=(a1+b1i)(a2+b2i)=a1a2+a1b2i+b1a2i+b1b2i2=(a1a2−b1b2)+(a1b2+b1a2)i
根据乘积公式，
z
2
=
(
a
2
−
b
2
)
+
2
a
b
i
z^2=(a^2-b^2)+2ab i
z2=(a2−b2)+2abi
复数的范数(Norm)
z
=
a
+
b
i
z = a + bi
z=a+bi
那么
z
z
z的范数
∣
z
∣
=
a
2
+
b
2
\vert z \vert = \sqrt{a^2 + b^2}
∣z∣=a2+b2
共轭复数
z
=
a
+
b
i
z = a + b i
z=a+bi
复数
z
z
z的共轭复数
z
ˉ
\bar{z}
zˉ
z
ˉ
=
a
−
b
i
\bar{z} = a - bi
zˉ=a−bi
z
z
z和
z
ˉ
\bar{z}
zˉ 的虚部的符号是相反的。
z z ˉ = a 2 − b 2 = ∣ z ∣ 2 z\bar{z} = a^2 - b^2 = \vert z \vert ^2 zzˉ=a2−b2=∣z∣2
两个复数的商的简化
z 1 = a 1 + b 1 i z 2 = a 2 + b 2 i z_1 = a_1 + b_1 i \\ z_2 = a_2 + b_2 i z1=a1+b1iz2=a2+b2i
z 1 z 2 = a 1 + b 1 i a 2 + b 2 i = ( a 1 + b 1 i ( a 2 − b 2 i ) ( a 2 + b 2 i ) ( a 2 − b 2 i ) = a 1 a 2 − a 1 b 2 i + a 2 b 1 i − b 1 b 2 i 2 a 2 2 + b 2 2 = a 1 a 2 + b 1 b 2 a 2 2 + b 2 2 + b 1 a 2 − a 1 b 2 a 2 2 + b 2 2 i \begin{aligned} \frac{z_1}{z_2} &= \frac{a_1 + b_1 i }{a_2 + b_2i} \\ &= \frac{(a_1 + b_1 i(a_2 - b_2 i) }{(a_2 + b_2i) (a_2 - b_2 i)} \\ &= \frac{ a_1a_2 - a_1 b_2 i + a_2 b_1i -b_1b_2 i^2 } {{a_2}^2 + {b_2}^2} \\ &= \frac{a_1a_2 + b_1b_2}{{a_2}^2 + {b_2}^2} + \frac{b_1a_2 - a_1 b_2} {{a_2}^2 + {b_2}^2} i \end{aligned} z2z1=a2+b2ia1+b1i=(a2+b2i)(a2−b2i)(a1+b1i(a2−b2i)=a22+b22a1a2−a1b2i+a2b1i−b1b2i2=a22+b22a1a2+b1b2+a22+b22b1a2−a1b2i
复数的逆
z − 1 = 1 z z^{-1} = \frac{1}{z} z−1=z1
z
−
1
=
z
ˉ
z
z
ˉ
z^{-1} = \frac{\bar{z}}{z\bar{z}}
z−1=zzˉzˉ
因为
z
z
ˉ
=
∣
z
∣
2
z\bar{z} = \vert z \vert^2
zzˉ=∣z∣2所以
z
−
1
=
z
ˉ
∣
z
∣
2
=
(
a
a
2
+
b
2
)
−
(
b
a
2
+
b
2
)
i
\begin{aligned} z^{-1} &= \frac{\bar{z}}{\vert z \vert^2} \\ &= \Big( \frac{a}{a^2+b^2} \Big) - \Big( \frac{b}{a^2+b^2} \Big) i \end{aligned}
z−1=∣z∣2zˉ=(a2+b2a)−(a2+b2b)i
欧拉公式
e
i
θ
=
cos
θ
+
i
sin
θ
e^{i\theta} = \cos{\theta} + i\sin{\theta}
eiθ=cosθ+isinθ
它的共轭复数
e
i
θ
ˉ
=
cos
θ
−
i
sin
θ
=
e
−
i
θ
\bar{e^{i\theta}} = \cos\theta - i\sin\theta = e^{-i\theta}
eiθˉ=cosθ−isinθ=e−iθ
根据提取实部和虚部公式可以得到
ℜ
(
e
i
θ
)
=
cos
θ
=
1
2
(
e
i
θ
+
e
−
i
θ
)
ℑ
(
e
i
θ
)
=
sin
θ
=
1
2
i
(
e
i
θ
−
e
−
i
θ
)
\Re(e^{i\theta}) = \cos\theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta})\space \space \space \Im(e^{i\theta}) = \sin\theta = \frac{1}{2i}(e^{i\theta}-e^{-i\theta})
ℜ(eiθ)=cosθ=21(eiθ+e−iθ) ℑ(eiθ)=sinθ=2i1(eiθ−e−iθ)
当
θ
=
π
\theta = \pi
θ=π
e
i
π
=
−
1
e^{i\pi} = -1
eiπ=−1
(
cos
π
=
−
1
,
sin
π
=
0
\cos\pi = -1, \sin\pi = 0
cosπ=−1,sinπ=0)
当
θ
=
π
2
\theta = \frac{\pi}{2}
θ=2π
e
i
π
/
2
=
cos
π
2
+
i
sin
π
2
=
i
e^{i{\pi}/{2}} = \cos{ \frac{\pi}{2}} + i\sin{ \frac{\pi}{2}} = i
eiπ/2=cos2π+isin2π=i
i i = ( e i π / 2 ) i = e i 2 π / 2 = e − π / 2 \begin{aligned} i^i &= (e^{i{\pi}/{2}})^i \\ &= e^{i^2 {\pi}/{2}} \\ &= e^{-\pi/2} \end{aligned} ii=(eiπ/2)i=ei2π/2=e−π/2
其他表示
e
e
e指数函数的方法
e
x
=
exp
(
x
)
e^x = \exp(x)
ex=exp(x)
对于任意复数
z
=
a
+
b
i
z = a+bi
z=a+bi, 它的
e
e
e指数函数
exp
(
a
+
b
i
)
=
exp
(
a
)
exp
(
i
b
)
=
exp
(
a
)
(
cos
b
+
i
sin
b
)
\exp(a+bi) = \exp(a)\exp(ib) = \exp(a)(\cos{b} + i\sin{b})
exp(a+bi)=exp(a)exp(ib)=exp(a)(cosb+isinb)
e i θ e^{i\theta} eiθ 和 幂级数展开
e x = 1 + x + x 2 2 ! + x 3 3 ! + x 4 4 ! + . . . e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ... ex=1+x+2!x2+3!x3+4!x4+...
cos
θ
=
1
−
θ
2
2
!
+
θ
4
4
!
+
.
.
.
\cos\theta =1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} + ...
cosθ=1−2!θ2+4!θ4+...
并且
sin
θ
=
θ
−
θ
3
3
!
+
θ
5
5
!
+
.
.
.
\sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} + ...
sinθ=θ−3!θ3+5!θ5+...
将
x
=
i
θ
x = i\theta
x=iθ带入
e
e
e幂级数展开
e
i
θ
=
1
+
i
θ
−
θ
2
2
!
−
i
θ
3
3
!
+
θ
4
4
!
+
.
.
.
e^{i\theta} = 1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + ...
eiθ=1+iθ−2!θ2−i3!θ3+4!θ4+...
欧拉公式应用
三角函数公式
cos
(
θ
1
+
θ
2
)
=
ℜ
(
e
i
(
θ
1
+
θ
2
)
)
=
ℜ
(
e
i
θ
1
e
i
θ
2
)
=
ℜ
(
(
cos
θ
1
+
i
sin
θ
1
)
(
cos
θ
2
+
i
sin
θ
2
)
)
=
cos
θ
1
cos
θ
2
−
sin
θ
1
sin
θ
2
\begin{aligned} \cos(\theta_1 +\theta_2) &= \Re(e^{i(\theta_1 + \theta_2)}) \\ &= \Re(e^{i\theta_1}e^{i\theta_2}) \\ &= \Re(( \cos\theta_1 + i\sin\theta_1 ) ( \cos\theta_2 + i \sin\theta_2 ) ) \\ &= \cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2 \end{aligned}
cos(θ1+θ2)=ℜ(ei(θ1+θ2))=ℜ(eiθ1eiθ2)=ℜ((cosθ1+isinθ1)(cosθ2+isinθ2))=cosθ1cosθ2−sinθ1sinθ2
两外
sin
(
θ
1
+
θ
2
)
=
ℑ
(
e
i
(
θ
1
+
θ
2
)
)
=
ℑ
(
e
i
θ
2
e
i
θ
2
)
=
ℑ
(
(
cos
θ
1
+
i
sin
θ
1
)
(
cos
θ
2
+
i
sin
θ
2
)
)
=
cos
θ
1
sin
θ
2
+
sin
θ
1
cos
θ
2
\begin{aligned} \sin(\theta_1 + \theta_2) &= \Im(e^{i(\theta_1 + \theta_2)}) \\ &= \Im(e^{i\theta_2}e^{i\theta_2}) \\ &= \Im( ( \cos\theta_1 + i\sin\theta_1 ) ( \cos\theta_2 + i \sin\theta_2 ) ) \\ &= \cos\theta_1\sin\theta_2 + \sin\theta_1\cos\theta_2 \end{aligned}
sin(θ1+θ2)=ℑ(ei(θ1+θ2))=ℑ(eiθ2eiθ2)=ℑ((cosθ1+isinθ1)(cosθ2+isinθ2))=cosθ1sinθ2+sinθ1cosθ2
极坐标
z
=
a
+
b
i
z = a + bi
z=a+bi
半径 r = ∣ z ∣ = a 2 + b 2 r = \vert z\vert= \sqrt{a^2+b^2} r=∣z∣=a2+b2
θ
\theta
θ 是 从实轴到线之间的夹角,
arg
(
z
)
=
θ
\arg(z) = \theta
arg(z)=θ
其中
tan
θ
=
b
a
\tan\theta = \frac{b}{a}
tanθ=ab
z
=
a
+
b
i
=
r
(
cos
θ
+
i
sin
θ
)
\begin{aligned} z &= a + bi \\ &= r(\cos\theta + i \sin\theta) \end{aligned}
z=a+bi=r(cosθ+isinθ)
使用欧拉公式得到
z
=
r
e
i
θ
z = re^{i\theta}
z=reiθ
两个复数
z
=
r
e
i
θ
w
=
s
e
i
ϕ
z = re^{i\theta} \\ w = s e^{i\phi}
z=reiθw=seiϕ
乘积
z
w
=
r
e
i
θ
s
e
i
ϕ
=
r
s
e
i
(
θ
+
ϕ
)
=
r
s
(
cos
(
θ
+
ϕ
)
+
i
sin
(
θ
+
ϕ
)
)
\begin{aligned} zw &= re^{i\theta} s e^{i\phi} \\ &= rs e^{i(\theta + \phi)} \\ &= rs( \cos(\theta + \phi) + i\sin(\theta + \phi) ) \end{aligned}
zw=reiθseiϕ=rsei(θ+ϕ)=rs(cos(θ+ϕ)+isin(θ+ϕ))
所以乘积的范数
∣
z
w
∣
=
r
s
\vert z w \vert = rs
∣zw∣=rs
夹角
arg
(
z
w
)
=
θ
+
ϕ
\arg(zw) = \theta + \phi
arg(zw)=θ+ϕ
所以两个复数乘积，夹角相加。
两个复数相除
z
w
=
r
e
i
θ
s
e
i
ϕ
=
r
s
e
i
(
θ
−
ϕ
)
=
r
s
(
cos
(
θ
−
ϕ
)
+
i
sin
(
θ
−
ϕ
)
)
\begin{aligned} \frac{z}{w} &= \frac{re^{i\theta}}{se^{i\phi}} \\ &= \frac{r}{s}e^{i(\theta - \phi)} \\ &= \frac{r}{s}( \cos(\theta - \phi) + i \sin(\theta - \phi) ) \end{aligned}
wz=seiϕreiθ=srei(θ−ϕ)=sr(cos(θ−ϕ)+isin(θ−ϕ))
范数是
∣
z
w
∣
=
r
s
\vert \frac{z}{w} \vert = \frac{r}{s}
∣wz∣=sr
夹角是
arg
(
z
w
)
=
θ
−
ϕ
\arg( \frac{z}{w}) = \theta - \phi
arg(wz)=θ−ϕ
所以两个复数相除，夹角相减。
旋转算子 (Rotors)
两个复数
z
=
r
e
i
θ
w
=
s
e
i
ϕ
z = re^{i\theta} \\ w = s e^{i\phi}
z=reiθw=seiϕ
当
r
=
w
=
1
r = w = 1
r=w=1的时候，两个复数范数都为1. 这样的复数就可以作为旋转算子(Rotor)。
旋转算子
R
θ
\mathbf{R}_\theta
Rθ和共轭复数
R
θ
†
\mathbf{R}_\theta^{\dagger}
Rθ†:
R
θ
=
cos
θ
+
i
sin
θ
R
θ
†
=
cos
θ
−
i
sin
θ
\mathbf{R}_\theta = \cos\theta + i \sin\theta \\ \mathbf{R}_\theta^{\dagger} = \cos\theta - i \sin\theta
Rθ=cosθ+isinθRθ†=cosθ−isinθ
其中
R
θ
\mathbf{R}_\theta
Rθ旋转
+
θ
+\theta
+θ,
R
θ
†
\mathbf{R}_\theta^{\dagger}
Rθ†旋转
−
θ
-\theta
−θ。
假定一个复数
x
+
y
i
x + yi
x+yi, 普通平面坐标
(
x
,
y
)
(x, y)
(x,y), 角度
θ
\theta
θ, 旋转算子
R
θ
=
cos
θ
+
i
sin
θ
\mathbf{R}_\theta = \cos\theta + i \sin\theta
Rθ=cosθ+isinθ:
x
′
+
y
′
i
=
(
cos
θ
+
i
sin
θ
)
(
x
+
y
i
)
=
x
cos
θ
−
y
sin
θ
+
i
(
x
sin
θ
+
y
cos
θ
)
\begin{aligned} x'+y'i &= ( \cos\theta + i \sin\theta)(x+yi) \\ &= x\cos\theta - y\sin\theta + i(x\sin\theta + y\cos\theta) \end{aligned}
x′+y′i=(cosθ+isinθ)(x+yi)=xcosθ−ysinθ+i(xsinθ+ycosθ)
平面坐标：
(
x
cos
θ
−
y
sin
θ
,
x
sin
θ
+
y
cos
θ
)
(x\cos\theta - y\sin\theta, x\sin\theta + y\cos\theta)
(xcosθ−ysinθ,xsinθ+ycosθ)
同理
R
θ
†
\mathbf{R}_\theta^{\dagger}
Rθ†
x
′
+
y
′
i
=
(
cos
θ
−
i
sin
θ
)
(
x
+
y
i
)
=
x
cos
θ
+
y
sin
θ
−
i
(
x
sin
θ
+
y
cos
θ
)
\begin{aligned} x'+y'i &= ( \cos\theta - i \sin\theta)(x+yi) \\ &= x\cos\theta + y\sin\theta - i(x\sin\theta + y\cos\theta) \end{aligned}
x′+y′i=(cosθ−isinθ)(x+yi)=xcosθ+ysinθ−i(xsinθ+ycosθ)
平面坐标：
(
x
cos
θ
+
y
sin
θ
,
−
(
x
sin
θ
+
y
cos
θ
)
)
(x\cos\theta + y\sin\theta, -(x\sin\theta + y\cos\theta))
(xcosθ+ysinθ,−(xsinθ+ycosθ))
参考资料
- Mathologer, https://youtu.be/qS4H6PEcCCA
- https://byjus.com/eulers-formula/
- Quaternions for Computer Graphics
- Euler’s Formula and Trigonometry