反三角函数的的相互关系

arcsin⁡x=−arcsin⁡(−x)=π2−arccos⁡x=arctan⁡x1−x2=arccos⁡1−x2=arccot⁡1−x2x(1)\begin{align} \arcsin x&=-\arcsin(-x)\\ &=\frac\pi 2-\arccos x\\&=\arctan\frac{x}{\sqrt{1-x^2}} \\&=\arccos\sqrt{1-x^2} \\ &=\operatorname{arccot}\frac{\sqrt{1-x^2}}x \end{align}\tag 1 arcsinx​=−arcsin(−x)=2π​−arccosx=arctan1−x2​x​=arccos1−x2​=arccotx1−x2​​​(1)

最后两个等号只在 x>0 时成立，下同

arccos⁡x=π−arccos⁡(−x)=π2−arcsin⁡x=arccot⁡x1−x2=arcsin⁡1−x2=arctan⁡1−x2x(2)\begin{align} \arccos x&=\pi-\arccos(-x)\\ &=\frac\pi 2-\arcsin x\\ &=\operatorname{arccot}\frac{x}{\sqrt{1-x^2}}\\ &=\arcsin\sqrt{1-x^2}\\ &=\arctan\frac{\sqrt{1-x^2}}x \end{align}\tag2 arccosx​=π−arccos(−x)=2π​−arcsinx=arccot1−x2​x​=arcsin1−x2​=arctanx1−x2​​​(2)

arctan⁡x=−arctan⁡(−x)=π2−arccot⁡x=arcsin⁡x1+x2=arccos⁡11+x2=arccot⁡1x(3)\begin{align} \arctan x&=-\arctan(-x)\\ &=\frac\pi 2-\operatorname{arccot}x\\ &=\arcsin\frac{x}{\sqrt{1+x^2}}\\ &=\arccos\frac 1{\sqrt{1+x^2}}\\ &=\operatorname{arccot}\frac 1x \end{align}\tag 3 arctanx​=−arctan(−x)=2π​−arccotx=arcsin1+x2​x​=arccos1+x2​1​=arccotx1​​(3)

arccot⁡x=π−arccot⁡(−x)=π2−arctan⁡x=arccos⁡x1+x2=arcsin⁡11+x2=arctan⁡1x(4)\begin{align} \operatorname{arccot}x&=\pi-\operatorname{arccot}(-x)\\&=\frac\pi2-\arctan x\\&=\arccos \frac x{\sqrt{1+x^2}}\\&=\arcsin\frac1{\sqrt{1+x^2}}\\&=\arctan\frac 1x\\ \end{align}\tag 4 arccotx​=π−arccot(−x)=2π​−arctanx=arccos1+x2​x​=arcsin1+x2​1​=arctanx1​​(4)

反三角函数的和差

反正弦：

arcsin⁡x+arcsin⁡y=arcsin⁡(x1−y2+y1−x2)(xy≤0orx2+y2≤1)=π−arcsin⁡(x1−y2+y1−x2)(x>0,y>0,x2+y2>1)=−π−arcsin⁡(x1−y2+y1−x2)(x0,y>0,x^2+y^2>1)\\ &=-\pi-\arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})\quad(x0,y>0,x2+y2>1)=−π−arcsin(x1−y2​+y1−x2​)(x0,y1)=−π−arcsin⁡(x1−y2−y1−x2)(x0,x2+y2>1)(6)\begin{align} \arcsin x-\arcsin y&=\arcsin(x\sqrt{1-y^2}-y\sqrt{1-x^2})\quad(xy\ge0\quad or\quad x^2+y^2\le1)\\ &=\pi-\arcsin(x\sqrt{1-y^2}-y\sqrt{1-x^2})\quad(x>0,y1)\\ &=-\pi-\arcsin(x\sqrt{1-y^2}-y\sqrt{1-x^2})\quad(x0,x^2+y^2>1)\\ \end{align}\tag6 arcsinx−arcsiny​=arcsin(x1−y2​−y1−x2​)(xy≥0orx2+y2≤1)=π−arcsin(x1−y2​−y1−x2​)(x>0,y1)=−π−arcsin(x1−y2​−y1−x2​)(x0,x2+y2>1)​(6)

反余弦：

arccos⁡x+arccos⁡y=arccos⁡[xy−(1−x2)(1−y2)](x+y≥0)=2π−arccos⁡[xy−(1−x2)(1−y2)](x+y