三角函数倍角公式推导

1.sin(2α)=2sin(α)cos(α)

已知单位圆中有圆周三角形△ABC,圆周角为α,圆心为点O
$\therefore 圆心角=2?圆周角$
$\because \angle AOC=\alpha ,AB=2cos(\alpha ),AC=2sin(\alpha )$
$做AP\perp BC于点P$
$\therefore △ABP与△APO共用边AP$
$\because sin(\alpha )?AB=sin(2\alpha )$
$\because sin(2\alpha )=2sin(\alpha )cos(\alpha )$

2.$cos(2\alpha )=cos(\alpha {)}^2?sin(\alpha {)}^2$

$\therefore 圆心角=2?圆周角$
$\because \angle AOC=\alpha ,AB=2cos(\alpha ),AC=2sin(\alpha )$
$做AP\perp BC于点P$
$\because \angle PAC+\angle ACP=\angle ABC+\angle ACP$
$\because \angle PAC=\alpha$
$\because OP=1?2sin(\alpha )sin(\alpha )=cos(2\alpha )$
$\because cos(2\alpha )=1?2sin(\alpha {)}^2$
$\because cos(2\alpha )=cos(\alpha {)}^2?sin(\alpha {)}^2$

3.$tan(2\alpha )=\frac{2tan(\alpha )}{1?tan(\alpha {)}^2}$

由公式1,2推tan(2α)
$tan(2\alpha )=\frac{sin(2\alpha )}{cos(2\alpha )}$
$=\frac{2sin(\alpha )cos(\alpha )}{cos(\alpha {)}^2?sin(\alpha {)}^2}$
$=\frac{2tan(\alpha )}{1?tan(\alpha {)}^2}$

4.$cot(2\alpha )=\frac{cot(\alpha {)}^2?1}{2cot(\alpha )}$

由公式1,2推cot(2α)
$cot(2\alpha )=\frac{cos(2\alpha )}{sin(2\alpha )}$
$=\frac{cos(\alpha {)}^2?sin(\alpha {)}^2}{2sin(\alpha )cos(\alpha )}$
$=\frac{cot(\alpha {)}^2?1}{2cot(\alpha )}$