单位圆
弧度制
三角函数
$$\sin (x) = {对边 \over 斜边} , \cos (x) = {邻边 \over 斜边} , \tan (x) = {对边 \over 斜边}$$
$$\csc (x) = {1 \over \sin (x)} , \sec = {1 \over \cos (x) } , \cot (x) = {1 \over \tan (x)}$$
常用三角函数值
| 0 | $${\pi \over 6}$$ | $${\pi \over 4}$$ | $${\pi \over 3}$$ | $${\pi \over 2}$$ | |
|---|---|---|---|---|---|
| $$\sin$$ | $$0$$ | $${1 \over 2}$$ | $$1 \over \sqrt{2}$$ | $$\sqrt{3} \over 2$$ | $$1$$ |
| $$\cos$$ | $$1$$ | $$\sqrt{3} \over 2$$ | $$1 \over \sqrt{2}$$ | $${1 \over 2}$$ | $$0$$ |
| $$\tan$$ | $$0$$ | $$1 \over \sqrt{3}$$ | $$1$$ | $$\sqrt{3}$$ | $$*$$ |
ASTC
奇偶性
$$\sin(-x) = -\sin(x) , \tan(-x) = -\tan(x) , \cos(-x) = \cos(x)$$
三角恒等式
$$\tan(x) = {\sin(x) \over \cos(x)} , \cot(x) = {\cos(x) \over \sin(x)}$$
毕达哥拉斯定理
$$\sin^{2}(x) + \cos^{2}(x) = 1$$
$$1 + \tan^{2}(x) = \sec^{2}(x)$$
$$1 + \cot^{2}(x) = \csc^{2}(x)$$
两角和
$$\sin(A+B)= \sin(A)\cdot\cos(B) + \cos(A)\cdot\sin(B) , \sin(A-B)= \sin(A)\cdot\cos(B) - \cos(A)\cdot\sin(B)$$
$$\cos(A+B)= \cos(A)\cdot\cos(B) - \sin(A)\cdot\sin(B) , \cos(A-B)= \cos(A)\cdot\cos(B) + \sin(A)\cdot\sin(B)$$
倍角公式
$$\sin(2x) = 2\sin(x)\cos(x)$$
$$\cos(2x) = cos^{2}(x) - sin^{2}(x) = 2\cos^{2}(x) - 1 = 1 - 2\sin^{2}(x)$$
