Solve the Problem $C [ 0 , \pi ]$ Have the Inner Product

Solve the problem.
-Let $C [ 0 , \pi ]$ have the inner product $\langle f , g \rangle = \int _ { 0 } ^ { \pi } f ( t ) g ( t ) d t$ , and let $m$ and $n$ be unequal positive integers. Prove that $\cos ( \mathrm { mt } )$ and $\cos ( \mathrm { nt } )$ are orthogonal.

A) $\langle \cos ( m t ) , \cos ( n t ) \rangle = \int _ { 0 } ^ { \pi } \cos ( m t ) \cos ( n t ) d t$
$$\begin{array} { l } = \frac { 1 } { 2 } \int _ { 0 } ^ { \pi } [ \cos ( m t + n t ) + \cos ( m t - n t ) ] d t \\= \frac { 1 } { 2 } \left[ \frac { \sin ( m t + n t ) } { m - n } + \frac { \sin ( m t - n t ) } { m + n } \right] \text { from } [ 0 , \pi ] \\= 1 .\end{array}$$
B) $$\text { } \begin{aligned}\langle \cos ( m t ) , \cos ( n t ) \rangle & = \int _ { 0 } ^ { \pi } \cos ( m t ) \cos ( n t ) d t \\& = \int _ { 0 } ^ { \pi } [ \cos ( m t + n t ) + \cos ( m t - n t ) ] d t \\& = \left[ \frac { \sin ( m t - n t ) } { m + n } + \frac { \sin ( m t + n t ) } { m - n } \right] \text { from } [ 0 , \pi ] \\& = 1 .\end{aligned}$$
C) $$\text { } \begin{aligned}\langle \cos ( m t ) , \cos ( n t ) \rangle & = \int _ { 0 } ^ { \pi } \cos ( m t ) \cos ( n t ) d t \\& = \frac { 1 } { 2 } \int _ { 0 } ^ { \pi } [ \cos ( m t + n t ) + \cos ( m t - n t ) ] d t \\& = \frac { 1 } { 2 } \left[ \frac { \sin ( m t + n t ) } { m + n } + \frac { \sin ( m t - n t ) } { m - n } \right] \text { from } [ 0 , \pi ] \\& = 0 .\end{aligned}$$
D) $$\text {} \begin{aligned}\langle \cos ( m t ) , \cos ( n t ) \rangle & = \int _ { 0 } ^ { \pi } \cos ( m t ) \cos ( n t ) d t \\& = \frac { 1 } { 2 } \int _ { 0 } ^ { \pi } [ \cos ( m t + n t ) - \cos ( m t - n t ) ] d t \\& = \frac { 1 } { 2 } \left[ \frac { \sin ( m t + n t ) } { m - n } - \frac { \sin ( m t - n t ) } { m + n } \right] \text { from } [ 0 , \pi ]\end{aligned}$$