万能公式
$sin^2alpha + cos^2alpha = 1$
勾股定理
和角公式
$sin(alpha+eta) = sinalphacoseta + cosalphasineta$
$cos(alpha+eta) = cosalphacoseta - sinalphasineta$
$ an(alpha+eta) = dfrac{ analpha + aneta}{1 - analpha aneta}$
差角公式
$sin(alpha-eta) = sinalphacoseta - cosalphasineta$
$cos(alpha-eta) = cosalphacoseta + sinalphasineta$
$ an(alpha-eta) = dfrac{ analpha - aneta}{1 + analpha aneta}$
和角公式差角公式的推导
在单位圆中，用向量$overrightarrow{OA}$与向量$overrightarrow{OB}$分别代表角$alpha,eta$的终边，$x$轴正半轴为始边，则
$overrightarrow{OA} = (cosalpha, sinalpha), overrightarrow{OB} = (coseta, sineta)$
则 $overrightarrow{OA}·overrightarrow{OB} = cosalphacoseta + sinalphasineta$
设其夹角为$θ$，则$overrightarrow{OA}·overrightarrow{OB} = |overrightarrow{OA}|·|overrightarrow{OB}| cos(θ) = cosalphacoseta + sinalphasineta$
因此$cos(alpha-eta) = cosalphacoseta + sinalphasineta$
又因为$cos(alpha+eta) = cos(alpha-(-eta))$，因此有$cos(alpha+eta) = cosalphacos(-eta) + sinalphasin(-eta) = cosalphacoseta - sinalphasineta$
又因为诱导公式$sinalpha = cos(dfrac{π}{2}-alpha)$
因此$sin(alpha+eta) = cos(dfrac{π}{2}-alpha-eta) = cos(dfrac{π}{2}-alpha)coseta+sin(dfrac{π}{2}-alpha)sineta = sinalphacoseta + cosalphasineta$
同理可推得$sin(alpha-eta)$
$ an(alpha+eta) = dfrac{sin(alpha+eta)}{cos(alpha+eta)} = dfrac{sinalphacoseta - cosalphasineta}{cosalphacoseta + sinalphasineta}$
上下同时除以$cosalphacoseta$，即可得$ an(alpha+eta) = dfrac{ analpha + aneta}{1 - analpha aneta}$
同理可推得$ an(alpha-eta)$
和差化积公式 (一次同名)
$sinalpha + sineta = 2sin(dfrac{alpha+eta}{2})cos(dfrac{alpha-eta}{2})$
$sinalpha - sineta = 2cos(dfrac{alpha+eta}{2})sin(dfrac{alpha-eta}{2})$
$cosalpha + coseta = 2cos(dfrac{alpha+eta}{2})cos(dfrac{alpha-eta}{2})$
$cosalpha - coseta = 2sin(dfrac{alpha+eta}{2})sin(dfrac{alpha-eta}{2})$
$ analpha + aneta = dfrac{sin(alpha+eta)}{cosalphacoseta}$
$ analpha - aneta = dfrac{sin(alpha-eta)}{cosalphacoseta}$
和差化积公式的推导
$sinalpha = sin(dfrac{alpha+eta}{2} +dfrac{alpha-eta}{2} ) = sin(dfrac{alpha+eta}{2})cos(dfrac{alpha-eta}{2})+sin(dfrac{alpha-eta}{2})cos(dfrac{alpha+eta}{2})$
$sineta = sin(dfrac{alpha+eta}{2} -dfrac{alpha-eta}{2} ) = sin(dfrac{alpha+eta}{2})cos(dfrac{alpha-eta}{2})-sin(dfrac{alpha-eta}{2})cos(dfrac{alpha+eta}{2})$
两式相加即可得$sinalpha + sineta = 2sin(dfrac{alpha+eta}{2})cos(dfrac{alpha-eta}{2})$
同理可推导$cosalpha + coseta$与$cosalpha - coseta$
$ analpha + aneta = dfrac{sinalpha}{cosalpha} + dfrac{sineta}{coseta}$，通分即可
倍角公式 （这里是指二倍角公式）
$sin(2alpha) = 2sinalphacosalpha$
$cos(2alpha) = cos^2alpha-sin^2alpha$
$ an(2alpha) = dfrac{2 analpha}{1- an^2alpha}$
以上公式利用和角公式证明即可
由于$sin^2alpha + cos^2alpha = 1$，可得
$cos(2alpha) = 2cos^2alpha-1$
$cos(2alpha) = 1-2sin^2alpha$
半角公式
$sin(dfrac{alpha}{2}) = ±sqrt{dfrac{1-cosalpha}{2}}$
$cos(dfrac{alpha}{2}) = ±sqrt{dfrac{1+cosalpha}{2}}$
$ an(dfrac{alpha}{2}) = ±sqrt{dfrac{1-cosalpha}{1+cosalpha}}$
半角公式的推导
由倍角公式$cos(2alpha) = 1-2sin^2alpha$，将$alpha$替换为$dfrac{alpha}{2}$得$cosalpha = 1-2sin^2(dfrac{alpha}{2})$，即$2sin^2(dfrac{alpha}{2}) = 1-cosalpha$
同理利用倍角公式$cos(2alpha) = 2cos^2alpha-1$可推得$cos(dfrac{alpha}{2})$
$ an(dfrac{alpha}{2}) = dfrac{sin(dfrac{alpha}{2})}{cos(dfrac{alpha}{2})} = dfrac{±sqrt{dfrac{1-cosalpha}{2}}}{±sqrt{dfrac{1+cosalpha}{2}}} = ±sqrt{dfrac{1-cosalpha}{1+cosalpha}} $