Question

Show that for all integers m and n, with m ≠ +/-n, the integral from -π to π of cos(mθ)cos(nθ) dθ = 0

Answer 1

We have this equality :
\[\cos a\cos b=\frac12\left(\cos(a+b)+\cos(a-b)\right)\]
So :
\[\int_{-\pi}^\pi\cos(m\theta)\cos(n\theta)\mathrm d\theta=\int_{-\pi}^\pi\frac12\left(\cos((n+m)\theta)+\cos((n-m)\theta)\right)\mathrm d\theta
\\=\frac{1}{2}\left[\frac1{n+m}\sin((n+m)\theta)+\frac1{n-m}\sin((n-m)\theta)\right]_{-\pi}^\pi\\
\\=0
\]


Anyway, we can solve it faster by remarking that the function :
\[f(\theta)=\cos(n\theta)\cos(m\theta)\]
is a pair function , and therefore its integral over [-pi,pi] is zero !

Answer 2

thanks.
let b and c be integers and let p and q be distinct primes. suppose p^2 + bpq + cq^2 = 0. define a sequence a base n = n^2 + bn +c . show that a base n is not = 0 for all integers n >or = 1