求解方程A*cos + B*sin = C

問題

求解
$A *\cos(\theta) + B*\sin(\theta) = C$
中的$\theta$值。

方法一

採用三角函數的萬能公式進行求解，假設$t = \tan(\frac{\theta}{2})$
其中
$\frac{1}{\cos(\frac{\theta}{2})^2} = 1 + \tan(\frac{\theta}{2})^2$
則

$\cos(\theta) = \cos(\frac{\theta}{2})^2 - \sin(\frac{\theta}{2})^2 \\ \qquad\qquad =\cos(\frac{\theta}{2})^2(1 - \frac{\sin(\frac{\theta}{2})^2}{\cos(\frac{\theta}{2})^2}) \\ \qquad \quad =\frac{1}{\frac{1}{\cos(\frac{\theta}{2})^2}}(1 - \tan(\frac{\theta}{2})^2)\\ \qquad \qquad \qquad =\frac{1}{1 + \tan(\frac{\theta}{2})^2}(1 - \tan(\frac{\theta}{2})^2)\\ \!\!=\frac{1 - \tan(\frac{\theta}{2})^2}{1 + \tan(\frac{\theta}{2})^2}\\ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!=\frac{1 - t^2}{1 + t^2}$

$\sin(\theta) = 2*\sin(\frac{\theta}{2})*\cos(\frac{\theta}{2}) \\ \qquad\qquad= 2*(\frac{\sin(\frac{\theta}{2})}{\cos(\frac{\theta}{2})})*\cos(\frac{\theta}{2})^2\\ \qquad\qquad\quad= 2*\tan(\frac{\theta}{2})*\frac{1}{1+\tan(\frac{\theta}{2})^2}\\ = \frac{2*\tan(\frac{\theta}{2})}{1+\tan(\frac{\theta}{2})^2}\\ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!= \frac{2*t}{1 + t^2}$
則上述方程可以寫爲
$A*\frac{1 - t^2}{1 + t^2} + B*\frac{2*t}{1 + t^2} = C$
化簡得
$(A+C)*t^2-2*B*t-(A-C)=0$
故二次方程的解爲
$t = \frac{B\pm\sqrt{B^2 + (A + C)(A - C)}}{A + C}$
所以方程的解爲
$\theta = 2*\arctan(t)$