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Next: Mixed sine and cosine Up: Orthogonality of trigonometric functions Previous: Preliminary results

Pair of sines or cosines

From (1), we have

$\displaystyle \int_0^\pi \cos mx \cos nx \d x = \int_0^\pi \sin mx \sin nx \d x$

Substituting this into (2) gives

$\displaystyle 2 \int_0^\pi \cos mx \cos nx  \d x = 2 \int_0^\pi \sin mx \sin nx
 \d x = 0
$

Hence, for $ n\neq m$, we have

$\displaystyle \int_0^\pi \cos mx \cos nx  \d x = 0
$

and

$\displaystyle \int_0^\pi \sin mx \sin nx  \d x = 0
$

Alternatively, we can prove the same results by integrating by parts, as follows.
$\displaystyle K$ $\displaystyle :=$ $\displaystyle \int_0^\pi \cos mx \cos nx \d x
$  
  $\displaystyle =$ $\displaystyle \frac{1}{m} \biggl[ \sin mx \cos nx \biggr]_0^\pi + \frac{n}{m}
\int_0^\pi \sin mx \sin nx  \d x$  
  $\displaystyle =$ $\displaystyle \frac{n}{m} K$  

(by (1)). Hence, $ K=0$. This argument also holds only for $ n\neq m$, since otherwise $ nK/m$ becomes $ K$ and so we get $ K=K$, so that $ K$ is not necessarily zero.

In the case when $ n=m$, we have

$\displaystyle \int_0^\pi \cos mx \cos nx \d x
$ $\displaystyle =$ $\displaystyle \int_0^\pi \cos^2 nx  \d x$  
  $\displaystyle =$ $\displaystyle \frac{1}{2} \int_0^\pi \biggl( 1 + \cos 2nx \biggr)  \d x$  
  $\displaystyle =$ $\displaystyle \frac{1}{2} \biggl[ x + \frac{1}{2n} \sin 2nx \biggr]_0^\pi$  
  $\displaystyle =$ $\displaystyle \frac{1}{2} \pi$  

Since (1) is valid for $ n=m$ as well as $ n\neq m$, we have

$\displaystyle \int_0^\pi \cos mx \cos nx  \d x = \int_0^\pi \sin mx \sin nx  \d x
= \frac{1}{2} \pi \delta_{mn}
$

where $ \delta_{mn}$ is the Kronecker delta.
next up previous
Next: Mixed sine and cosine Up: Orthogonality of trigonometric functions Previous: Preliminary results
Alexander Frolkin 2001-02-17