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将$$$\left(3 \sqrt{2}, - \frac{\sqrt{2}}{4}\right)$$$绕$$$\left(0, 0\right)$$$逆时针旋转$$$45^{\circ}$$$
您的输入
将$$$\left(3 \sqrt{2}, - \frac{\sqrt{2}}{4}\right)$$$绕$$$\left(0, 0\right)$$$逆时针旋转$$$45^{\circ}$$$角。
解答
将点$$$\left(x, y\right)$$$绕原点按角度$$$\theta$$$逆时针旋转，会得到新点$$$\left(x \cos{\left(\theta \right)} - y \sin{\left(\theta \right)}, x \sin{\left(\theta \right)} + y \cos{\left(\theta \right)}\right)$$$。
在我们的情况下，$$$x = 3 \sqrt{2}$$$、$$$y = - \frac{\sqrt{2}}{4}$$$ 和 $$$\theta = 45^{\circ}$$$。
因此，新点是 $$$\left(3 \sqrt{2} \cos{\left(45^{\circ} \right)} - - \frac{\sqrt{2}}{4} \sin{\left(45^{\circ} \right)}, 3 \sqrt{2} \sin{\left(45^{\circ} \right)} + - \frac{\sqrt{2}}{4} \cos{\left(45^{\circ} \right)}\right) = \left(\frac{13}{4}, \frac{11}{4}\right)$$$。
答案
新点为 $$$\left(\frac{13}{4}, \frac{11}{4}\right) = \left(3.25, 2.75\right)$$$A。