Show that $\{1, \cos x, \sin x, \cos(2x), \sin(2x), \dots, \cos(nx), \sin(nx)\}$ is an orthogonal set of V .

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Suppose $V := CR([0, 2\pi])$ is the collection of real valued functions on $[0, 2\pi]$ with inner product given by $$\langle f,g\rangle:= \int_0^{2\pi}f(x)g(x)\,dx$$

Show that $$S_n := \{1, \cos x, \sin x, \cos(2x), \sin(2x), \dots , \cos(nx), \sin(nx)\}$$ is an orthogonal set of V.

Find the orthogonal projection of a function $f \in V$ on $span(S_n).$ Particularly find the orthogonal projections of the function $x$ on $span(S_2)$ and $span(S_3).$

(And we know for $n \to \infty,$ the orthogonal projection of a function $f$ on $S_n$ is nothing but the Fourier series expansion of the function $f$).

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$$\cos (a x) \sin (b x)=\frac{1}{2} (\sin (a x+b x)-\sin (a x-b x))$$ Integrating $$\int_0^{2\pi} \cos (a x) \sin (b x)\,dx=\int_0^{2\pi} \frac{1}{2} (\sin (a x+b x)-\sin (a x-b x))\,dx=$$ $$=\left[\frac{\cos (x (a-b))}{2 (a-b)}-\frac{\cos (x (a+b))}{2 (a+b)}\right]_0^{2\pi}=\\=\frac{\cos (2 \pi (a-b))}{2 (a-b)}-\frac{\cos (2 \pi (a+b))}{2 (a+b)}-\left(\frac{1}{2 (a-b)}-\frac{1}{2 (a+b)}\right)=0,\;\forall a,b\in\mathbb{Z}$$ Edit

For the other products $\sin(ax)\sin(bx);\;\cos(ax)\cos(bx)$ a similar procedure leads to the same result.