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Trigonometric Orthogonality
Orthogonality of trigonometric functions is a very important mathematical property, especially in signal processing and Fourier analysis. Orthogonality of trigonometric functions can be expressed as that the integral of the product of sine and cosine functions of different frequencies is equal to zero within one period.
集合 { sin 0 x = 0 , cos 0 x = 1 , sin x , cos x , sin 2 x , cos 2 x , … } \{\sin 0x = 0, \cos 0x = 1, \sin x, \cos x, \sin 2x, \cos 2x, \ldots\} { sin0x _=0,cos0x _=1,sinx,cosx,sin2 x ,cos2 x ,. . . } contains an infinite number of sine and cosine functions with increasing frequency starting from 0.
These functions are in [ 0 , 2 π ] [0, 2\pi][0,2 π ] is orthogonal. This means that if you take any two of these distinct functions and compute the integral of their product over a full cycle, the result will be zero.
Specifically:
- when n = mn = mn=m时，∫ 0 2 π sin nx sin mx dx = ∫ 0 2 π cos nx cos mx dx = π \int_{0}^{2\pi} \sin nx \sin mx \,dx = \ int_{0}^{2\pi} \cos nx \cos mx \,dx = \pi∫02 p.msinnxsinmxdx=∫02 p.mcosnxcosmxdx=p ;
- 当n ≠ mn \neq mn=m 时， ∫ 0 2 π sin n x sin m x d x = ∫ 0 2 π cos n x cos m x d x = 0 \int_{0}^{2\pi} \sin nx \sin mx \,dx = \int_{0}^{2\pi} \cos nx \cos mx \,dx = 0 ∫02 p.msinnxsinmxdx=∫02 p.mcosnxcosmxdx=0；
- For all n , mn, mn,m， ∫ 0 2 π sin n x cos m x d x = 0 \int_{0}^{2\pi} \sin nx \cos mx \,dx = 0 ∫02 p.msinnxcosmxdx=0。
Orthogonality of the sine function
Orthogonality of the sine function depends on the integer nnn andmmwhether m is equal. Below we demonstrate these two cases in detail.
1. When n = mn = mn=when m :
Consider points:
∫ 0 2 π sin nx sin nx dx = ∫ 0 2 π sin 2 nx dx \int_{0}^{2\pi} \sin nx \sin nx \,dx = \int_{0}^{2 \pi} \sin^2 nx \,dx∫02 p.msinnxsinnxdx=∫02 p.msin2nxdx
We can convert this to cosine form using the trigonometric identity:
sin 2 n x = 1 − cos 2 n x 2 \sin^2 nx = \frac{1 - \cos 2nx}{2} sin2nx=21−cos2nx
Points get:
∫ 0 2 π 1 − cos 2 nx 2 dx = π \int_{0}^{2\pi} \frac{1 - \cos 2nx}{2} \,dx = \pi∫02 p.m21−cos2nxdx=Pi
-
Complete derivation, we can get:
∫ 0 2 π sin nx sin nx dx = ∫ 0 2 π sin 2 nx dx = ∫ 0 2 π 1 − cos 2 nx 2 dx = 1 2 ∫ 0 2 π 1 dx − 1 2 ∫ 0 2 π cos 2 nx dx = 1 2 ⋅ 2 π − 0 = π \begin{aligned}\int_{0}^{2\pi} \sin nx \sin nx \,dx &= \int_{0}^{2 \pi} \sin^2nx\,dx\\&= \int_{0}^{2\pi}\frac{1 - \cos 2nx}{2}\,dx\\&= \frac{1} {2}\int_{0}^{2\pi} 1 \,dx - \frac{1}{2}\int_{0}^{2\pi} \cos 2nx \,dx \\&= \frac {1}{2}\cdot 2\pi - 0 \\&= \pi\end{aligned}∫02 p.msinnxsinnxdx=∫02 p.msin2nxdx=∫02 p.m21−cos2nxdx=21∫02 p.m1dx−21∫02 p.mcos2nxdx=21⋅2 p.m−0=p
2. 当n ≠ mn \neq mn=when m :
Consider points:
∫ 0 2 π sin n x sin m x d x \int_{0}^{2\pi} \sin nx \sin mx \,dx ∫02 p.msinnxsinmxdx
We can use the product-to-sum identity:
sin n x sin m x = 1 2 ( cos ( n − m ) x − cos ( n + m ) x ) \sin nx \sin mx = \frac{1}{2}\left( \cos(n-m)x - \cos(n+m)x \right) sinnxsinmx=21(cos(n−m)x−cos(n+m)x)
然后积分:
∫ 0 2 π sin n x sin m x d x = 1 2 ∫ 0 2 π ( cos ( n − m ) x − cos ( n + m ) x ) d x = 1 2 [ sin ( n − m ) x n − m − sin ( n + m ) x n + m ] 0 2 π = 1 2 [ sin ( n − m ) ( 2 π ) n − m − sin ( n + m ) ( 2 π ) n + m − sin ( n − m ) ( 0 ) n − m + sin ( n + m ) ( 0 ) n + m ] = 0 \begin{aligned}\int_{0}^{2\pi} \sin nx \sin mx \,dx &= \frac{1}{2}\int_{0}^{2\pi} \left( \cos(n-m)x - \cos(n+m)x\right) \,dx \\ &= \frac{1}{2}\left[\frac{\sin(n-m)x}{n-m} - \frac{\sin(n+m)x}{n+m}\right]_{0}^{2\pi} \\ &= \frac{1}{2}\left[\frac{\sin(n-m)(2\pi)}{n-m} - \frac{\sin(n+m)(2\pi)}{n+m} - \frac{\sin(n-m)(0)}{n-m} + \frac{\sin(n+m)(0)}{n+m}\right] \\ &= 0\end{aligned} ∫02 p.msinnxsinmxdx=21∫02 p.m(cos(n−m)x−cos(n+m)x)dx=21[n−mwithout ( no−m)x−n+mwithout ( no+m)x]02 p.m=21[n−mwithout ( no−m)(2π)−n+mwithout ( no+m)(2π)−n−mwithout ( no−m)(0)+n+mwithout ( no+m)(0)]=0
Because ( n − m ) (nm)(n−m ) and( n + m ) (n+m)(n+m ) are all non-zero integers, so in[ 0 , 2 π ] [0, 2\pi][0,2 π ] on the cosine function integral is 0.
Summarize:
- when n = mn = mn=When m , the integral result isπ \pip .
- 当n ≠ mn \neq mn=m , the integral result is 0.
These properties are very important in Fourier analysis and signal processing, and they allow us to decompose complex periodic signals into sine and cosine components of different frequencies.
Orthogonality of cosine function
The orthogonality of the cosine function can also be proved by a similar method. We analyze two cases:
1. When n = mn = mn=when m :
Consider points:
∫ 0 2 π cos n x cos n x d x = ∫ 0 2 π cos 2 n x d x \int_{0}^{2\pi} \cos nx \cos nx \,dx = \int_{0}^{2\pi} \cos^2 nx \,dx ∫02 p.mcosnxcosnxdx=∫02 p.mcos2nxdx
We can convert this to cosine form using the trigonometric identity:
cos 2 n x = 1 + cos 2 n x 2 \cos^2 nx = \frac{1 + \cos 2nx}{2} cos2nx=21+cos2nx
Points get:
∫ 0 2 π 1 + cos 2 nx 2 dx = π \int_{0}^{2\pi} \frac{1 + \cos 2nx}{2} \,dx = \pi∫02 p.m21+cos2nxdx=Pi
-
Complete derivation, we can get:
∫ 0 2 π cos nx cos nx dx = ∫ 0 2 π cos 2 nx dx = ∫ 0 2 π 1 + cos 2 nx 2 dx = 1 2 ∫ 0 2 π 1 dx + 1 2 ∫ 0 2 π cos 2 nx dx = 1 2 ⋅ 2 π + 0 = π \begin{aligned}\int_{0}^{2\pi} \cos nx \cos nx \,dx &= \int_{0}^{2 \pi} \cos^2nx\,dx\\&= \int_{0}^{2\pi}\frac{1 + \cos2nx}{2}\,dx\\&= \frac{1} {2}\int_{0}^{2\pi} 1 \,dx + \frac{1}{2}\int_{0}^{2\pi} \cos 2nx \,dx \\&= \frac {1}{2}\cdot 2\pi + 0 \\&= \pi\end{aligned}∫02 p.mcosnxcosnxdx=∫02 p.mcos2nxdx=∫02 p.m21+cos2nxdx=21∫02 p.m1dx+21∫02 p.mcos2nxdx=21⋅2 p.m+0=p
2. 当n ≠ mn \neq mn=when m :
Consider points:
∫ 0 2 π cos n x cos m x d x \int_{0}^{2\pi} \cos nx \cos mx \,dx ∫02 p.mcosnxcosmxdx
We can use the product-to-sum identity:
cos n x cos m x = 1 2 ( cos ( n − m ) x + cos ( n + m ) x ) \cos nx \cos mx = \frac{1}{2}\left( \cos(n-m)x + \cos(n+m)x \right) cosnxcosmx=21(cos(n−m)x+cos(n+m)x)
然后积分:
∫ 0 2 π cos n x cos m x d x = 1 2 ∫ 0 2 π ( cos ( n − m ) x + cos ( n + m ) x ) d x = 1 2 [ sin ( n − m ) x n − m + sin ( n + m ) x n + m ] 0 2 π = 1 2 [ sin ( n − m ) ( 2 π ) n − m + sin ( n + m ) ( 2 π ) n + m − sin ( n − m ) ( 0 ) n − m − sin ( n + m ) ( 0 ) n + m ] = 0 \begin{aligned}\int_{0}^{2\pi} \cos nx \cos mx \,dx &= \frac{1}{2}\int_{0}^{2\pi} \left( \cos(n-m)x + \cos(n+m)x\right) \,dx \\&= \frac{1}{2}\left[\frac{\sin(n-m)x}{n-m} + \frac{\sin(n+m)x}{n+m}\right]_{0}^{2\pi} \\&= \frac{1}{2}\left[\frac{\sin(n-m)(2\pi)}{n-m} + \frac{\sin(n+m)(2\pi)}{n+m} - \frac{\sin(n-m)(0)}{n-m} -\frac{\sin(n+m)(0)}{n+m}\right] \\&= 0\end{aligned} ∫02 p.mcosnxcosmxdx=21∫02 p.m(cos(n−m)x+cos(n+m)x)dx=21[n−mwithout ( no−m)x+n+mwithout ( no+m)x]02 p.m=21[n−mwithout ( no−m)(2π)+n+mwithout ( no+m)(2π)−n−mwithout ( no−m)(0)−n+mwithout ( no+m)(0)]=0
Because ( n − m ) (nm)(n−m ) and( n + m ) (n+m)(n+m ) are all non-zero integers, so in[ 0 , 2 π ] [0, 2\pi][0,2 π ] on the cosine function integral is 0.
Summarize:
- when n = mn = mn=When m , the integral result isπ \pip .
- 当n ≠ mn \neq mn=m , the integral result is 0.
These properties are equally important in Fourier analysis and signal processing, where they also allow us to decompose complex periodic signals into cosine components of different frequencies.
Orthogonality of the sine and cosine functions
The orthogonality of the sine and cosine functions can also be proved, we analyze two cases:
when n = mn = mn=when m :
Consider points:
∫ 0 2 π sin n x cos n x d x \int_{0}^{2\pi} \sin nx \cos nx \,dx ∫02 p.msinnxcosnxdx
Using the product-to-sum identity, we have:
sin n x cos n x = 1 2 ( sin ( n + m ) x + sin ( n − m ) x ) \sin nx \cos nx = \frac{1}{2}\left( \sin(n + m)x + \sin(n - m)x \right) sinnxcosnx=21( without ( no+m)x+without ( no−m)x)
Since n = mn = mn=m , the expression becomes:
1 2 ( sin 2 n x + sin 0 ) \frac{1}{2}\left( \sin 2nx + \sin 0 \right) 21(sin2nx+sin0)
Points get:
∫ 0 2 π 1 2 ( sin 2 n x + sin 0 ) d x = 0 \int_{0}^{2\pi} \frac{1}{2}\left( \sin 2nx + \sin 0 \right) \,dx = 0 ∫02 p.m21(sin2nx+sin0)dx=0
当n ≠ mn \neq mn=when m :
Consider points:
∫ 0 2 π sin n x cos m x d x \int_{0}^{2\pi} \sin nx \cos mx \,dx ∫02 p.msinnxcosmxdx
We can use the product-to-sum identity:
sin n x cos m x = 1 2 ( sin ( n + m ) x + sin ( n − m ) x ) \sin nx \cos mx = \frac{1}{2}\left( \sin(n + m)x + \sin(n - m)x \right) sinnxcosmx=21( without ( no+m)x+without ( no−m)x)
Then points:
∫ 0 2 π 1 2 ( sin ( n + m ) x + sin ( n − m ) x ) d x = 0 \int_{0}^{2\pi} \frac{1}{2}\left( \sin(n + m)x + \sin(n - m)x \right) \,dx = 0 ∫02 p.m21( without ( no+m)x+without ( no−m)x)dx=0
Because ( n − m ) (nm)(n−m ) and( n + m ) (n+m)(n+m ) are all integers, so in[ 0 , 2 π ] [0, 2\pi][0,2 π ] on the integral of the sine function is 0.
- for any integer nn andmmm , we have:
∫ 0 2 π sin n x cos m x d x = 0 \int_{0}^{2\pi} \sin nx \cos mx \,dx = 0 ∫02 p.msinnxcosmxdx=0
The proof is as follows:
Use trigonometric identities:
sin n x cos m x = 1 2 ( sin ( n + m ) x + sin ( n − m ) x ) \sin nx \cos mx = \frac{1}{2} \left( \sin(n+m)x + \sin(n-m)x \right) sinnxcosmx=21( without ( no+m)x+without ( no−m)x)
We can get:
∫ 0 2 π sin n x cos m x d x = 1 2 ∫ 0 2 π ( sin ( n + m ) x + sin ( n − m ) x ) d x = 1 2 [ − cos ( n + m ) x n + m − cos ( n − m ) x n − m ] 0 2 π = 1 2 [ − cos ( n + m ) ( 2 π ) n + m + cos ( n + m ) ( 0 ) n + m + cos ( n − m ) ( 2 π ) n − m − cos ( n − m ) ( 0 ) n − m ] = 0 \begin{aligned} \int_0^{2\pi}\sin nx\cos mxdx& =\frac12\int_0^{2\pi}\left(\sin(n+m)x+\sin(n-m)x\right)dx \\ &=\frac12\left[-\frac{\cos(n+m)x}{n+m}-\frac{\cos(n-m)x}{n-m}\right]_0^{2\pi} \\ &=\frac12\left[-\frac{\cos(n+m)(2\pi)}{n+m}+\frac{\cos(n+m)(0)}{n+m}+\frac{\cos(n-m)(2\pi)}{n-m}-\frac{\cos(n-m)(0)}{n-m}\right] \\ &=0 \end{aligned} ∫02 p.msinnxcosmxdx=21∫02 p.m( without ( no+m)x+without ( no−m)x)dx=21[−n+mcos(n+m)x−n−mcos(n−m)x]02 p.m=21[−n+mcos(n+m)(2π)+n+mcos(n+m)(0)+n−mcos(n−m)(2π)−n−mcos(n−m)(0)]=0
In this way, we prove the orthogonality between the sine and cosine functions. When the frequencies of the two functions are different, the integral of the product over one period is equal to zero.
Summarize:
- when n = mn = mn=m , the integral result is 0.
- 当n ≠ mn \neq mn=m , the integral result is 0.
These properties are equally important in Fourier analysis, where they show that sine and cosine functions of different frequencies are orthogonal to each other over the entire period.