Given
Two signals
\begin{align} s_1(t) = \sin(2\pi ft) \end{align}
and
\begin{align} s_2(t) = \cos(2\pi ft) \end{align}
are defined for $t \in [0 T)$
Question
The task is to show that $s_1(t)$ and $s_2(t)$ are approximately orthogonal if $f >> 1/T$
Attempted solution
For orthogonality we must have that the inner product is equal to zero, that is
\begin{align} \langle s_1, s_2 \rangle = \int_0^T s_1(t)s_2(t)dt = 0 \end{align}
explicitly we have
\begin{align} \int_0^T s_1(t)s_2(t)dt = \int_0^T \sin(2\pi ft)\cos(2\pi ft)dt \end{align}
Using the trigonometric identity $\sin2\theta = 2\sin\theta\cos\theta$ we have
\begin{align} \int_0^T \frac{1}{2}\sin (4\pi ft)dt = \bigg [\frac{-cos(4\pi ft)}{8\pi f}\bigg ]_0^T = -\frac{1}{8\pi f}(\cos(4\pi fT) -1) \end{align}
If we set this expression to 0 we must have that $\cos(4\pi ft) - 1 = 0$, or
\begin{align} \cos(4\pi fT) = 1 \end{align}
or
\begin{align} 4\pi fT = 2\pi \end{align}
or
\begin{align} f = \frac{1}{2T} \end{align}
This doesn't make sense with the given condition, I mean why should $f$ be much bigger than $1/T$?
What you've shown is that if $f=1/(2T)$, then the two functions are orthogonal. However, what you're asked to do is show that if the frequency is far greater than $1/T$, the functions are approximately orthogonal. I would think the Riemann-Lebesgue Lemma might be useful here. You could probably take the real part of a complex exponential to get what you need.