จาก $\sum_{n = 1}^{44}\sin(45^{\circ}-n)=\sum_{n = 1}^{44}(\sin 45^{\circ}\cos n - \cos 45^{\circ}\sin n)=\dfrac{\sqrt{2}}{2}\sum_{n = 1}^{44}(\cos n - \sin n)$
เนื่องจาก $\sum_{n = 1}^{44}\sin(45^{\circ}-n)=\sum_{n = 1}^{44}\sin n$
ดังนั้น $\sum_{n = 1}^{44}\sin n=(\sqrt{2}-1)\sum_{n = 1}^{44}\cos n$
จากโจทย์ $\dfrac{\sum_{n = 1}^{44}cos(n) }{\sum_{n = 1}^{44}sin(n) }-\dfrac{\sum_{n = 1}^{44}sin(n) }{\sum_{n = 1}^{44}cos(n) }=\dfrac{\sum_{n = 1}^{44}cos(n) }{(\sqrt{2}-1)\sum_{n = 1}^{44}cos(n) }-\dfrac{(\sqrt{2}-1)\sum_{n = 1}^{44}cos(n) }{\sum_{n = 1}^{44}cos(n) }=\dfrac{1}{\sqrt{2}-1}-(\sqrt{2}-1) =2$
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