Why the right term can take kronecker delta ($\leftrightarrow$ non-zero) of even function and odd function of fourier transform?

Question

I've been encountering the problem of the below equation.

$$ \int_{-T_{0}/2 }^{T_{0}/2 } \cos\left(\frac{2 \pi m}{T_{0}} t \right) \cos\left(\frac{2 \pi n }{T_{0}} t\right) \,dt = \frac{T_{0} }{2} \delta_{m,n} $$

$$ m,n \in\mathbb{N} $$

I want to prove that the LHS of the equation can be non-zero.

The range of the ingration of LHS is symmetric against y-axis and since even function * even function= even function holds, the LHS of the equation must be always zero.

What have I been missing?

Answer 1

Use the formula $cos(x) cos(y) = 0.5(cos((x+y))+cos(x-y))$ https://byjus.com/maths/trigonometry-formulas/

$\int cos(nx) cos(mx) dx = 0.5 \int (cos((n-m)x) + cos((n+m)x)) dx = 0$ if $n \neq m$.