$$\cos (\theta)+i \sin (\theta)=e^{i \theta}$$

$$\sin (\theta)=\frac{e^{i \theta}-e^{-i \theta}}{2 i}$$

$$A=e^{\frac{i \pi}{180}} A^{90}=e^{\frac{i \pi}{2}}=i$$

\begin{aligned} & \text { 原式 }=\sum_{n=1}^{90} \sin \frac{n \pi}{180}=\frac{1}{2 i} \sum_{n=1}^{90}\left(e^{\frac{i n \pi}{180}}-e^{\frac{-i n \pi}{180}}\right)=\frac{1}{2 i} \sum_{n=1}^{90}\left(A^n-A^{-n}\right) \\ & =\frac{1}{2 i}\left[A \frac{1-A^{90}}{1-A}-\left(A^{-1}\right) \frac{1-A^{-90}}{1-A^{-1}}\right]=\frac{1}{2 i}\left[A \frac{1-i}{1-A}-\left(A^{-1}\right) \frac{1+i}{1-A^{-1}}\right] \\ & =\frac{1}{2 i}\left(\frac{A-i A}{1-A}-\frac{1+i}{A-1}\right)=\frac{1}{2 i} \frac{(1+A)+i(1-A)}{1-A} \\ & =\frac{1}{2 i} \frac{1+A}{1-A}+\frac{1}{2} \end{aligned}

\begin{aligned} & \frac{1}{2 i} \frac{1+A}{1-A}=\frac{1}{2 i} \frac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=\frac{1}{2 i} \frac{(1+\cos \theta+i \sin \theta)(1-\cos \theta+i \sin \theta)}{(1+\cos \theta)^2+\sin ^2 \theta} \\ & =\frac{\sin \theta}{(1-\cos \theta)^2+\sin ^2 \theta} \end{aligned}

$$\sin 1^{\circ}+\sin 2^{\circ}+\sin 3^{\circ} \ldots+\sin 90^{\circ}=\frac{1}{2}+\frac{\sin 1^{\circ}}{\left(1-\cos 1^{\circ}\right)^2+\sin ^2 1^{\circ}} \approx 57.794325064654814$$

$$1=\int_0^{\frac{\pi}{2}} \sin x d x \approx \sum_{n=1}^{90}\left(\sin \frac{n \pi}{180}\right) \cdot \frac{\pi}{180}$$

$$\sum_{n=1}^{90}\left(\sin \frac{n \pi}{180}\right) \approx \frac{180}{\pi}$$
①点击 首页查看更多试卷和试题 , 点击查看 本题所在试卷