$\phi =\left \{ \cos\frac{2 n \pi t}{T},\sin\frac{2 n \pi t}{T}: \forall n,m \in \mathbb{Z}>0 \right \}$ in L$[0,T]$

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MY problem : Show that the real signals of continuous time, defined in $L [0, T]$, with $T ∈ R> 0$, which form the set

$$\phi =\left \{ \cos\frac{2 n \pi t}{T},\sin\frac{2 n \pi t}{T}: \forall n,m \in \mathbb{Z}>0 \right \}$$

they are orthogonal, precisely in the time interval $t ∈ [0, T]$, under the internal product between real signals defined by

$$\left \langle f(t),g(t)) \right \rangle=\int_{0}^{T} f(t)g(t)dt$$

Well, when I resolve the integral don't get zero :c

I do not understand that

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Hint:

To calculate the integral $\:\displaystyle\int_0^T\cos\frac{2m\pi t}{T}\sin\frac{2n\pi t}{T}\,\mathrm dt$, use the linearisation formula:

$$ \sin a\cos b=\tfrac12\bigl(\sin(a+b)+\sin(a-b)\bigr). $$