ว่างแล้ว มาต่อกันสัก 2 ข้อ ดีกว่า
และข้อ 29 จัดให้ตามคำเรียกร้องงงงง... 
$\begin{array}{rcl} \sum_{n = 1}^{44} cos \ n^o & = & cos \ 1^o+cos \ 2^o+...+cos \ 43^o+cos \ 44^o \\
& = & (cos \ 1^o+cos \ 44^o)+(cos \ 2^o+cos \ 43^o)+...+(cos \ 22^o+cos \ 23^o)\\
& = & (2cos \ 22.5^ocos \ 21.5^o)+(2cos \ 22.5^ocos \ 20.5^o)+...+(2cos \ 22.5^ocos \ 0.5^o)\\
& = & 2cos \ 22.5^o(cos \ 21.5^o+cos \ 20.5^o+...+cos \ 0.5^o) ---- (1) \end{array}$
$\begin{array}{rcl} \sum_{n = 1}^{44} sin \ n^o & = & sin \ 1^o+sin \ 2^o+...+sin \ 43^o+sin \ 44^o \\
& = & (sin \ 1^o+sin \ 44^o)+(sin \ 2^o+sin \ 43^o)+...+(sin \ 22^o+sin \ 23^o)\\
& = & (2sin \ 22.5^ocos \ 21.5^o)+(2sin \ 22.5^ocos \ 20.5^o)+...+(2sin \ 22.5^ocos \ 0.5^o)\\
& = & 2sin \ 22.5^o(cos \ 21.5^o+cos \ 20.5^o+...+cos \ 0.5^o) ---- (2) \end{array}$
ให้ $a = \dfrac {\sum_{n = 1}^{44} sin \ n^o} {\sum_{n = 1}^{44} cos \ n^o } = \dfrac {สมการ (2)}{สมการ (1)} = \dfrac {sin \ 22.5^o2}{cos \ 22.5^o} = tan \ 22.5^o $
$\begin{array}{rcl} \dfrac {\sum_{n = 1}^{44} cos \ n^o} {\sum_{n = 1}^{44} sin \ n^o } - \dfrac {\sum_{n = 1}^{44} sin \ n^o} {\sum_{n = 1}^{44} cos \ n^o } & = & \dfrac{1}{a}-a = \dfrac{1}{tan \ 22.5^o}-tan \ 22.5^o\\ & = & \dfrac {1-tan^2 \ 22.5^o}{tan \ 22.5^o} = 2(\dfrac {1-tan^2 \ 22.5^o}{2tan \ 22.5^o}) \\ & = & \dfrac{2}{tan \ 45^o} = 2 \ Ans \end{array} $