Question

Find the orthogonal projection of f=sin x onto g=cos x

Answer 1

Answer and Explanation:

Consider the curves{eq}C[a,b]. {/eq}

{eq}\left \langle f,g \right \rangle= \int_{a}^{b} f(x)\: g(x) dx.\\ C[-\pi,\pi],\ \: \: f(x)=\sin x, \ g(x)=\cos x \\ \left \langle \sin x,\cos x \right \rangle= \int_{-\pi}^{\pi} \sin x\:\cos x \: dx.=\left ( \frac{1}{2}\sin ^2\left(x\right) \right )_{-\pi}^{\pi}=0{/eq}

Therefore, the given vectors are orthogonal

{eq}\left \langle \cos x,\cos x \right \rangle= \int_{-\pi}^{\pi} \cos x\:\cos x \: dx.=\left (\frac{1}{2}\left(x+\frac{1}{2}\sin \left(2x\right)\right) \right )_{-\pi}^{\pi}=\pi{/eq}

The orthogonal projection of f onto g={eq}\frac{\left \langle f,g \right \rangle}{\left \langle g,g \right \rangle}g=0{/eq