三角函数

特殊角的三角函数

$sin(\frac{1}{6}\pi)=\frac{1}{2}$
$sin(\frac{1}{3}\pi)=\frac{\sqrt{3}}{2}$
$cos(\frac{1}{6}\pi)=\frac{\sqrt{3}}{2}$
$cos(\frac{1}{3}\pi)=\frac{1}{2}$
$sin(\frac{1}{2}\pi)=\frac{\sqrt{2}}{2}$
$cos(\frac{1}{2}\pi)=\frac{\sqrt{2}}{2}$
$tan(\frac{1}{6}\pi)=\frac{\sqrt{3}}{3}$
$tan(\frac{1}{3}\pi)={\sqrt{3}}$
$tan({\frac{1}{4}\pi})={1}$

和差角公式

$sin(a+b)=sin(a)*cos(b)+cos(a)*sin(b)$

$sin(a-b)=sin(a)*cos(b)-cos(a)*sin(b)$

$cos(a+b)=cos(a)*cos(b)-sin(a)*sin(b)$

$cos(a-b)=cos(a)*cos(b)+sin(a)*sin(b)$

$tan(a+b)=\frac{tan(a)+tan(b)}{1-tan(a)*tan(b)}$

$tan(a-b)=\frac{tan(a)-tan(b)}{1+tan(a)*tan(b)}$

二倍角公式

$cos(2a)=cos(a)^{2}-sin(a)^{2}=2*cos(a)^{2}-1=1-2*sin(a)^{2}$

$sin(2a)=2*sin(a)*cos(a)$

$tan(2a)=\frac{2*tan(a)}{1-tan(a)^{2}}$

万能公式

$tan(2a)=\frac{2*tan(a)}{1-tan(a)^{2}}$

$cos(2a)=\frac{cos(a)^{2}-sin(a)^{2}}{cos(a)^{2}+sin(a)^{2}}=\frac{1-tan(a)^{2}}{1+tan(a)^{2}}$

$sin(2a)=\frac{2*sin(a)*cos(a)}{cos(a)^{2}+sin(a)^{2}}=\frac{2*tan(a)}{1+tan(a)^{2}}$

辅助角公式

$a*sin(q)+b*cos(q)=\sqrt{a^{2}+b^{2}}*(\frac{a}{\sqrt{a^{2}+b^{2}}}*sin(q)+\frac{a}{\sqrt{a^{2}+b^{2}}}*cos(q))$
令$\frac{a}{\sqrt{a^{2}+b^{2}}}=cos(j),\frac{b}{\sqrt{a^{2}+b^{2}}}=sin(j)$
原式=
$\sqrt{a^{2}+b^{2}}*(sin(q)*cos(j)+cos(q)*sin(j))=\sqrt{a^{2}+b^{2}}*sin(q+j)$