棣美弗公式推導三角函數倍角、半角公式

高中數學教材的棣美弗公式 $(cos\theta+i \,sin\theta)^n=cos\, n\theta+i sin\, n\theta$，n是整數。

倍角公式

$cos\, 2\theta+i sin\, 2\theta=(cos\theta+i \,sin\theta)^2=$
$cos^2\theta+2(cos\theta)(i \,sin\theta)+(i \,sin\theta)^2=$
$cos^2\theta+2isin\theta cos\theta-sin^2\theta=(cos^2\theta-sin^2\theta)+2isin\theta cos\theta$

得$$cos2\theta=cos^2\theta-sin^2\theta$$

$$sin2\theta=2sin\theta cos\theta$$

$cos\, 3\theta+i sin\, 3\theta=(cos\theta+i \,sin\theta)^3=$

$(cos\theta)^3+3 cos\theta(isin\theta)^2+3( cos\theta)^2(isin\theta)+(i\,sin\theta)^3=$

$cos^3\theta-3 cos\theta sin^2\theta+3 cos^2\theta (isin\theta)-i\,sin^3\theta=$

$cos^3\theta-3 cos\theta (1-cos^2\theta)+3(1- sin^2\theta) (isin\theta)-i\,sin^3\theta=$

$(4cos^3\theta-3 cos\theta) +i(3 sin\theta-4sin^3\theta)=$

得$$cos3\theta=4cos^3\theta-3 cos\theta$$

$$sin3\theta=3 sin\theta-4sin^3\theta$$

半角公式

$cos\theta+i sin\theta=(cos\frac{\theta}{2}+i \,sin\frac{\theta}{2})^2=$

$(cos\frac{\theta}{2})^2+2(cos\frac{\theta}{2})(i \,sin\frac{\theta}{2}))+(i\,sin\frac{\theta}{2})^2=$

$(cos\frac{\theta}{2})^2+2(cos\frac{\theta}{2})(i \,sin\frac{\theta}{2}))-(sin\frac{\theta}{2})^2=$

$cos^2\frac{\theta}{2}-(1-cos^2\frac{\theta}{2})+i\,2(cos\frac{\theta}{2})(sin\frac{\theta}{2})=$$(2cos^2\frac{\theta}{2}-1)+i\,2(cos\frac{\theta}{2})(sin\frac{\theta}{2}))$

得 $cos\theta=2cos^2\frac{\theta}{2}-1$，因此

$$cos\frac{\theta}{2}=\pm\sqrt\frac{1+cos\theta }{2}$$

$cos\theta+i sin\theta=(cos\frac{\theta}{2}+i \,sin\frac{\theta}{2})^2=$

$(cos\frac{\theta}{2})^2+2(cos\frac{\theta}{2})(i \,sin\frac{\theta}{2}))-(sin\frac{\theta}{2})^2=$

$(1-sin^2\frac{\theta}{2})+2(cos\frac{\theta}{2})(i \,sin\frac{\theta}{2})-(sin^2\frac{\theta}{2})=$

$(1-2sin^2\frac{\theta}{2})+i\,2(cos\frac{\theta}{2})(sin\frac{\theta}{2}))=$

得 $cos\theta=1-2sin^2\frac{\theta}{2}$，因此

$$sin\frac{\theta}{2}=\pm\sqrt\frac{1-cos\theta }{2}$$=