三角函数公式整理

基本概念：角的定义、终边、单位圆…

$$sin A=frac{y}{r}$$ $$cos A=frac{x}{r}$$ $$tan A=frac{y}{x}$$ $$cot A=frac{x}{y}$$ $$sin^2 alpha + cos ^2 alpha=1$$ $$tan alpha=frac{sin alpha}{cos alpha}$$ $$cot alpha= frac{1}{tan alpha}$$

$sin^2alpha+cos^2alpha=1$

$tanalpha=frac{sinalpha}{cosalpha}$

$cos(alpha+beta)=cosalphacosbeta-sinalphasinbeta$

$cos(alpha-beta)=cosalphacosbeta+sinalphasinbeta$

$sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta$

$sin(alpha-beta)=sinalphacosbeta-cosalphasinbeta$

$tan(alpha+beta)=frac{tanalpha+tanbeta}{1-tanalphatanbeta}$

$tan(alpha-beta)=frac{tanalpha-tanbeta}{1+tanalphatanbeta}$

$$asinalpha+bcosalpha=sqrt{a^2+b^2}sin(alpha+theta)$$ 其中$tantheta=frac{b}{a}.$

$sin2alpha=2sinalphacosalpha$

$cos2alpha=cos^2alpha-sin^2alpha=2cos^2alpha-1=1-2sin^2alpha$

$tan2alpha=frac{2tanalpha}{1-tan^2alpha}$

$sin3alpha=3sinalpha-4sin^3alpha=4sin(frac{pi}{3}-alpha)sinalphasin(frac{pi}{3}+alpha)$

$cos3alpha=-3cosalpha+4cos^3alpha=4cos(frac{pi}{3}-alpha)cosalphacos(frac{pi}{3}+alpha)$

$tan3alpha=tan(frac{pi}{3}-alpha)tanalphatan(frac{pi}{3}+alpha)$

$sinfrac{alpha}{2}=pmsqrt{frac{1-cosalpha}{2}}$

$cosfrac{alpha}{2}=pmsqrt{frac{1+cosalpha}{2}}$

$tanfrac{alpha}{2}=pmsqrt{frac{1-cosalpha}{1+cosalpha}}=frac{1-cosalpha}{sinalpha}=frac{sinalpha}{1+cosalpha}$

$sinalpha=frac{2tan{frac{alpha}{2}}}{1+tan^2frac{alpha}{2}}$

$cosalpha=frac{1-tan^2frac{alpha}{2}}{1+tan^2frac{alpha}{2}}$

$tanalpha=frac{2tan{frac{alpha}{2}}}{1-tan^2frac{alpha}{2}}$

$sinalpha+sinbeta=2sin{frac{alpha+beta}{2}}cos{frac{alpha-beta}{2}}$

$sinalpha-sinbeta=2cos{frac{alpha+beta}{2}}sin{frac{alpha-beta}{2}}$

$cosalpha+cosbeta=2cos{frac{alpha+beta}{2}}cos{frac{alpha-beta}{2}}$

$cosalpha-cosbeta=-2sin{frac{alpha+beta}{2}}sin{frac{alpha-beta}{2}}$

$sinalphacosbeta=frac{1}{2}(sin(alpha+beta)+sin(alpha-beta))$

$cosalphasinbeta=frac{1}{2}(sin(alpha+beta)-sin(alpha-beta))$

$cosalphacosbeta=frac{1}{2}(cos(alpha+beta)+cos(alpha-beta))$

$sinalphasinbeta=-frac{1}{2}(cos(alpha+beta)-cos(alpha-beta))$

$1+cot^2alpha=csc^2alpha=frac{1}{sin^2alpha}$

$1+tan^2alpha=sec^2alpha=frac{1}{cos^2alpha}$

$$sin^2alpha=frac{1-cos2alpha}{2}$$ $$cos^2alpha=frac{1+cos2alpha}{2}$$