Question

The angle between two nonzero elementsvandwof an inner product space is defined as

$\mathbf{\measuredangle }\left(v,w\right)\mathbf{=}\mathit{a}\mathit{r}\mathit{c}\mathit{c}\mathit{o}\mathit{s}\frac{<v,w>}{||v||\mathbf{.}||w||}$

In the space $\left[-\pi ,\pi \right]$with inner product

$<f,g>\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}\underset{\mathbf{-}\mathbf{\pi }}{\overset{\mathbf{\pi }}{\mathbf{\int }}}\mathit{f}\left(t\right)\mathit{g}\left(t\right)\mathit{d}\mathit{t}$

find the angle between $\mathit{f}\left(t\right)\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\left(t\right)$and $\mathit{g}\left(t\right)\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\left(t+\delta \right)$where $\mathbf{0}\mathbf{\le }\mathit{\delta }\mathbf{\le }\mathbf{\pi }$. Hint: Use the formula $\mathit{c}\mathit{o}\mathit{s}\left(t+\delta \right)\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathit{\delta }\mathbf{\right)}\mathbf{-}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathit{\delta }\mathbf{\right)}$.

Short answer

The angle between f and g is $\angle \left(f,g\right)=\delta$.

Step-by-step solution

Integral for angles.

Consider the following conditions of integral to find the angle between f and g.

It will check three integrals.

1. 
   $\begin{array}{rcl}\underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }}{\cos }^{2}tdt& =& \underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }}\frac{\cos \left(2t\right)+1}{2}dt\\ & =& \frac{1}{2}\times 2\times \underset{0}{\overset{\mathrm{\pi }}{\int }}\left(\cos \left(2t\right)+1\right)dt\\ & =& \mathrm{\pi }\end{array}$
2. 
   $\begin{array}{rcl}\underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }}{\cos }^{2}tdt& =& \underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }}\frac{1-\cos \left(2t\right)}{2}dt\\ & =& \frac{1}{2}\times 2\times \underset{0}{\overset{\mathrm{\pi }}{\int }}\left(1-\cos \left(2t\right)+1\right)dt\\ & =& \mathrm{\pi }\end{array}$
3. 
   $\begin{array}{rcl}\underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }}\sin t\cos tdt& =& \underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }}\frac{\sin 2t}{2}dt\\ & =& 0\end{array}$
   

Now, observe the following

Length of the functions f and g.

Also, the length of the functions f and g can be found as,

Therefore, the angle between f and g

Hence, the value of the angle between f and g will be

$\angle \left(f,g\right)=\delta$