Question

In Exercises 5-14, the space is \(C\left[ {0,2\pi } \right]\) with inner product (6).

7. Show that \({\left\| {\cos kt} \right\|^2} = \pi \) and \({\left\| {\sin kt} \right\|^2} = \pi \) for \(k > 0\).

Short answer

It is proved that\({\left\| {\cos kt} \right\|^2} = \pi {\rm{ and }}{\left\| {\sin kt} \right\|^2} = \pi ,\,\,\,\forall k > 0\).

Step-by-step solution

Inner Product

The Inner Productfor any two arbitrary functions is given by:

\(\left\langle {f,g} \right\rangle = \int_0^{2\pi } {f\left( t \right)g\left( t \right)dt} \)

Proving the statement

It is given that,\(k \in + {\rm I}\).

Using theinner productrule, we have:

\(\begin{array}{c}{\left\| {\cos kt} \right\|^2} = \int_0^{2\pi } {{{\cos }^2}ktdt} \\ = \frac{1}{2}\int_0^{2\pi } {\left\{ {1 + \cos 2kt} \right\}dt} \\ = \pi \end{array}\)

Similarly,

\(\begin{array}{c}{\left\| {\sin kt} \right\|^2} = \int_0^{2\pi } {{{\sin }^2}ktdt} \\ = \frac{1}{2}\int_0^{2\pi } {\left\{ {1 - \cos 2kt} \right\}dt} \\ = \pi \end{array}\)

Hence, \({\left\| {\cos kt} \right\|^2} = \pi {\rm{ and }}{\left\| {\sin kt} \right\|^2} = \pi ,\,\,\,\forall k > 0\).