$\displaystyle C_1\{f\}[n] = \sum_{k=0}^{N} c_k f[k]cos\Bigl[\frac{π}{N}nk\Bigl], \text{where}\
c_k =
\left\{
\begin{array}{l}
\frac{1}{2}, k≡0\ (mod\ N), \\1,\ \text{otherwise}\end{array}
\right. \tag{4}$$\displaystyle C_1\{f\}[n] = \sum_{k=0}^{N} c_k f[k]\\cos\Bigl[\frac{π}{N}nk\Bigl], \text{where}\
c_k =
\left\{
\begin{array}{l}
\tfrac12, \qquad &k (\mod N)=0, \\
1,\qquad &\text{otherwise} \\
\end{array}
\right. \tag{4}$
DST-I (N+1) points:
$$S_1\{f\}[n] = \sum_{k=0}^{N} f[k]sin\Bigl[\frac{π}{N}nk\Bigl]=\sum_{k=0}^{N} c_k f[k] sin\Bigl[\frac{π}{N}nk\Bigl] \tag{5}$$$$S_1\{f\}[n] = \sum_{k=0}^{N} f[k]\sin\Bigl[\frac{π}{N}nk\Bigl]=\sum_{k=0}^{N} c_k f[k] \sin\Bigl[\frac{π}{N}nk\Bigl] \tag{5}$$
$$C_2\{f\}[n] = \sum_{k=0}^{N} f[k]cos\Bigl[\frac{π}{N}\Bigl(k+\frac{1}{2}\Bigl)n\Bigl] \tag{6}$$$$C_2\{f\}[n] = \sum_{k=0}^{N} f[k]\cos\Bigl[\frac{π}{N}\Bigl(k+\tfrac12\Bigl)n\Bigl] \tag{6}$$
$$S_2\{f\}[n] = \sum_{k=0}^{N} f[k]sin\Bigl[\frac{π}{N}\Bigl(k+\frac{1}{2}\Bigl)n\Bigl] \tag{7}$$$$S_2\{f\}[n] = \sum_{k=0}^{N} f[k]\sin\Bigl[\frac{π}{N}\Bigl(k+\tfrac12\Bigl)n\Bigl] \tag{7}$$
Inverse DCT-I:
$$C_1^{-1} = \frac{2}{N}C_1 \tag{8}$$
Inverse DST-I:
$$S_1^{-1} = \frac{2}{N}S_1 \tag{9}$$
Inverse DCT-II (N+1) points:
$$C_2^{-1}\{F_{C_2}\}[k] = \frac{2}{N}\sum_{n=0}^{N} c_n F_{C_2}[n]cos\Bigl[\frac{π}{N}n\Bigl(k+\frac{1}{2}\Bigl)\Bigl] \tag{10}$$$$C_2^{-1}\{F_{C_2}\}[k] = \frac{2}{N}\sum_{n=0}^{N} c_n F_{C_2}[n]\cos\Bigl[\frac{π}{N}n\Bigl(k+\tfrac12\Bigl)\Bigl] \tag{10}$$
$C_2^{-1}\{C_2\{f\}\}[k] = f[k]$ except for $k=N$ there it is going to be $f[N-1]$.
Inverse DST-II (N+1) points:
$$C_2^{-1}\{F_{S_2}\}[k] = \frac{2}{N}\sum_{n=0}^{N} c_n F_{S_2}[n]sin\Bigl[\frac{π}{N}n\Bigl(k+\frac{1}{2}\Bigl)\Bigl] \tag{11}$$$$C_2^{-1}\{F_{S_2}\}[k] = \frac{2}{N}\sum_{n=0}^{N} c_n F_{S_2}[n]\sin\Bigl[\frac{π}{N}n\Bigl(k+\tfrac12\Bigl)\Bigl] \tag{11}$$
$S_2^{-1}\{S_2\{f\}\}[k] = f[k]$ except for $k=N$ there it is going to be $-f[N-1]$. Also, they are not interchangeable.
$$\sum_{m=0}^{2N} A[m] c_m cos\Bigl[\frac{π}{N}nm\Bigl] = \sum_{m=0}^{N} p_m(A[m] + A[2N-m]) c_m cos\Bigl[\frac{π}{N}nm\Bigl]$$$$\sum_{m=0}^{2N} A[m] c_m \cos\Bigl[\frac{π}{N}nm\Bigl] = \sum_{m=0}^{N} p_m(A[m] + A[2N-m]) c_m \cos\Bigl[\frac{π}{N}nm\Bigl]$$
$p_m = \frac{1}{2}$$p_m = \tfrac12$ if $m = N$ else $p_m = 1$
from the other hand we have
$$\sum_{m=0}^{2N} \frac{A[m]}{c_m} c_m cos\Bigl[\frac{π}{N}nm\Bigl] = \sum_{x=0}^{N}\sum_{y=0}^{N} f[x]g[y]cos\Bigl[\frac{π}{N}n(x+y)\Bigl]$$$$\sum_{m=0}^{2N} \frac{A[m]}{c_m} c_m \cos\Bigl[\frac{π}{N}nm\Bigl] = \sum_{x=0}^{N}\sum_{y=0}^{N} f[x]g[y]\cos\Bigl[\frac{π}{N}n(x+y)\Bigl]$$
$$\sum_{x=0}^{N}\sum_{y=0}^{N} f[x]g[y]cos\Bigl[\frac{π}{N}n(x+y)\Bigl] = \\\sum_{x=0}^{N}\sum_{y=0}^{N} f[x]g[y]\biggl(cos\Bigl[\frac{π}{N}nx\Bigl]cos\Bigl[\frac{π}{N}ny\Bigl] - sin\Bigl[\frac{π}{N}nx\Bigl]sin\Bigl[\frac{π}{N}ny\Bigl]\biggl) = \\ \biggl(\sum_{x=0}^{N} f[x]cos\Bigl[\frac{π}{N}nx\Bigl]\biggl)\cdot\biggl(\sum_{y=0}^{N} g[y]cos\Bigl[\frac{π}{N}ny\Bigl]\biggl) - \biggl(\sum_{x=0}^{N} f[x]sin \Bigl[\frac{π}{N}nx\Bigl]\biggl)\cdot\biggl(\sum_{y=0}^{N} g[y]sin\Bigl[\frac{π}{N}ny\Bigl]\biggl)$$$$\sum_{x=0}^{N}\sum_{y=0}^{N} f[x]g[y]\cos\Bigl[\frac{π}{N}n(x+y)\Bigl] = \\\sum_{x=0}^{N}\sum_{y=0}^{N} f[x]g[y]\biggl(\cos\Bigl[\frac{π}{N}nx\Bigl]\cos\Bigl[\frac{π}{N}ny\Bigl] - \sin\Bigl[\frac{π}{N}nx\Bigl]\sin\Bigl[\frac{π}{N}ny\Bigl]\biggl) = \\ \biggl(\sum_{x=0}^{N} f[x]\cos\Bigl[\frac{π}{N}nx\Bigl]\biggl)\cdot\biggl(\sum_{y=0}^{N} g[y]\cos\Bigl[\frac{π}{N}ny\Bigl]\biggl) - \biggl(\sum_{x=0}^{N} f[x]\sin \Bigl[\frac{π}{N}nx\Bigl]\biggl)\cdot\biggl(\sum_{y=0}^{N} g[y]\sin\Bigl[\frac{π}{N}ny\Bigl]\biggl)$$
$$\sum_{m=0}^{2N} A[m] cos\Bigl[\frac{π}{N}\Bigl(m+\frac{1}{2}\Bigl)n\Bigl] = \sum_{m=0}^{2N-2} A[m] cos\Bigl[\frac{π}{N}\Bigl(m+\frac{1}{2}\Bigl)n\Bigl] = \sum_{m=0}^{N-1} (A[m] + A[2N-1-m]) cos\Bigl[\frac{π}{N}\Bigl(m+\frac{1}{2}\Bigl)n\Bigl] $$$$\sum_{m=0}^{2N} A[m] \cos\Bigl[\frac{π}{N}\Bigl(m+\tfrac12\Bigl)n\Bigl] = \sum_{m=0}^{2N-2} A[m] \cos\Bigl[\frac{π}{N}\Bigl(m+\tfrac12\Bigl)n\Bigl] = \sum_{m=0}^{N-1} (A[m] + A[2N-1-m]) \cos\Bigl[\frac{π}{N}\Bigl(m+\tfrac12\Bigl)n\Bigl] $$
And the expression is going to be:
$$\sum_{m=0}^{2N} A[m] cos\Bigl[\frac{π}{N}\Bigl(m+\frac{1}{2}\Bigl)n\Bigl] =
\sum_{m=0}^{2N} A[m] \Bigl(cos\Bigl[\frac{πn}{2N}\Bigl]cos\Bigl[\frac{π}{N}kn\Bigl] - sin\Bigl[\frac{πn}{2N}\Bigl]sin\Bigl[\frac{π}{N}kn\Bigl]\Bigl) = \\
cos\Bigl[\frac{πn}{2N}\Bigl](F^{/}_{C_1}G^{/}_{C_1} - F_{S_1}G_{S_1}) - sin\Bigl[\frac{πn}{2N}\Bigl](F^{/}_{C_1}G_{S_1} + F_{S_1}G^{/}_{C_1}) = \\
=F^{/}_{C_1}(cos\Bigl[\frac{πn}{2N}\Bigl]G^{/}_{C_1} - sin\Bigl[\frac{πn}{2N}\Bigl]G_{S_1}) - F_{S_1}(cos\Bigl[\frac{πn}{2N}\Bigl]G_{S_1}+ sin\Bigl[\frac{πn}{2N}\Bigl]G^{/}_{C_1}) = \\
=F^{/}_{C_1}G_{C_2} - F_{S_1}G_{S_2} = G^{/}_{C_1}F_{C_2} - G_{S_1}F_{S_2} $$$$\sum_{m=0}^{2N} A[m] \cos\Bigl[\frac{π}{N}\Bigl(m+\tfrac12\Bigl)n\Bigl] =
\sum_{m=0}^{2N} A[m] \Bigl(\cos\Bigl[\frac{πn}{2N}\Bigl]\cos\Bigl[\frac{π}{N}kn\Bigl] - \sin\Bigl[\frac{πn}{2N}\Bigl]\sin\Bigl[\frac{π}{N}kn\Bigl]\Bigl) = \\
\cos\Bigl[\frac{πn}{2N}\Bigl](F^{/}_{C_1}G^{/}_{C_1} - F_{S_1}G_{S_1}) - \sin\Bigl[\frac{πn}{2N}\Bigl](F^{/}_{C_1}G_{S_1} + F_{S_1}G^{/}_{C_1}) = \\
=F^{/}_{C_1}(\cos\Bigl[\frac{πn}{2N}\Bigl]G^{/}_{C_1} - \sin\Bigl[\frac{πn}{2N}\Bigl]G_{S_1}) - F_{S_1}(\cos\Bigl[\frac{πn}{2N}\Bigl]G_{S_1}+ \sin\Bigl[\frac{πn}{2N}\Bigl]G^{/}_{C_1}) = \\
=F^{/}_{C_1}G_{C_2} - F_{S_1}G_{S_2} = G^{/}_{C_1}F_{C_2} - G_{S_1}F_{S_2} $$
