\\
\begin{array}{l}

(1) 第1象限(x:+,　y:+) \\
\begin{array}{ll}
　x&: \cos \theta \\
　y&: \sin \theta \\
　傾き(\frac{y}{x})&: \tan \theta (= \frac{\sin \theta}{\cos \theta}) \\
\end{array}
　\\
　\\
(2) 第1象限(x:+,　y:+) \\
\begin{array}{lll}
　x&: \cos(\frac{\pi}{2} - \theta) &= \sin \theta \\
　y&: \sin(\frac{\pi}{2} - \theta) &= \cos \theta \\
　傾き(\frac{y}{x})&: \tan(\frac{\pi}{2} - \theta) &= \frac{1}{\tan \theta} (= \frac{\cos\theta}{\sin\theta}) 
  \\
\end{array}
  \\
  \\
(2)' 第4象限(x:+,　y:-)　※(7)と同じ \\
\begin{array}{lll}
　x&: \cos(\theta - \frac{\pi}{2}) &= \sin \theta \\
　y&: \sin(\theta - \frac{\pi}{2}) &= - \cos \theta \\
　傾き(\frac{y}{x})&: \tan(\theta - \frac{\pi}{2}) &= - \frac{1}{\tan \theta} (= - \frac{\cos\theta}{\sin\theta}) 
\\
\end{array}
　\\
　\\
(3) 第2象限(x:-,　y:+) \\
\begin{array}{lll}
　x&: \cos(\frac{\pi}{2} + \theta) &= -\sin \theta \\
　y&: \sin(\frac{\pi}{2} + \theta) &= \cos \theta \\
　傾き(\frac{y}{x})&: \tan(\frac{\pi}{2} + \theta) &= -\frac{1}{\tan \theta} (= - \frac{\cos \theta}{\sin \theta}) \\
\end{array}
　\\
　\\
(4) 第2象限(x:-,　y:+) \\
\begin{array}{ll}
　x&: \cos({\pi} - \theta) &= -\cos \theta \\
　y&: \sin({\pi} - \theta) &= \sin \theta \\
　傾き(\frac{y}{x})&: \tan({\pi} - \theta) &= -\tan \theta (= - \frac{\sin\theta}{\cos \theta}) \\
\end{array}
  \\
  \\
(4)' 第3象限(x:-,　y:-)　※(5)と同じ \\
\begin{array}{ll}
　x&: \cos(\theta - {\pi}) &= - \cos \theta \\
　y&: \sin(\theta - {\pi}) &= - \sin \theta \\
　傾き(\frac{y}{x})&: \tan(\theta - {\pi}) &= \tan \theta (=  \frac{\sin\theta}{\cos \theta}) \\
\end{array}

\end{array}

\\
\begin{array}{l}

(5) 第3象限(x:-,　y:-) \\
\begin{array}{ll}
　x: \cos({\pi} + \theta) = - \cos \theta \\
　y: \sin({\pi} + \theta) = - \sin \theta \\
　傾き(\frac{y}{x}): \tan({\pi} + \theta) = \tan \theta (= \frac{\sin \theta}{\cos \theta}) \\
\end{array}

　\\
　\\
(6) 第3象限(x:-,　y:-) \\
\begin{array}{ll}
　x&: \cos(\frac{3}{2}\pi - \theta) &= - \sin \theta \\
　y&: \sin(\frac{3}{2}\pi - \theta) &= - \cos \theta \\
　傾き(\frac{y}{x})&: \tan(\frac{3}{2}\pi - \theta) &= \frac{1}{\tan \theta} (= \frac{\cos \theta}{\sin \theta}) \\
\end{array}
  \\
  \\
(6)' 第2象限(x:-,　y:+)　※(3)と同じ \\
\begin{array}{ll}
　x&: \cos(\theta - \frac{3}{2}\pi) &= - \sin \theta \\
　y&: \sin(\theta - \frac{3}{2}\pi) &= \cos \theta \\
　傾き(\frac{y}{x})&: \tan(\theta - \frac{3}{2}\pi) &= - \frac{1}{\tan \theta} (= - \frac{\cos \theta}{\sin \theta}) \\
\end{array}

　\\
　\\
(7) 第4象限(x:+,　y:-) \\
\begin{array}{ll}
　x&: \cos(\frac{3}{2}\pi + \theta) &= \sin \theta \\
　y&: \sin(\frac{3}{2}\pi + \theta) &= -\cos \theta \\
　傾き(\frac{y}{x})&: \tan(\frac{3}{2}\pi + \theta) &= - \frac{1}{\tan \theta} (= - \frac{\cos \theta}{\sin \theta}) \\
\end{array}

　\\
　\\
(8) 第4象限(x:+,　y:-) \\
\begin{array}{ll}
　x&: \cos(-\theta) &= \cos \theta \\
　y&: \sin(-\theta) &= -\sin \theta \\
　傾き(\frac{y}{x})&: \tan(-\theta) &= -\tan \theta (= - \frac{\sin \theta}{\cos \theta}) \\
\end{array}
　\\
　\\
電気関係で使うので、(3)(7)より、\\
-\cos(\theta + \frac{\pi}{2}) = \cos(\theta - \frac{\pi}{2}) = \sin \theta　※符号は \cos の位置で考える。\\
\sin(\theta + \frac{\pi}{2}) = -\sin(\theta - \frac{\pi}{2}) = \cos \theta　※符号は \sin の位置で考える。\\
\theta が第一象限にあるとすると、 \theta + \frac{\pi}{2}=第２象限、\theta - \frac{\pi}{2}=第４象限\\

\end{array}

\\
正弦定理 \\
\begin{array}{l}
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \\
\end{array}
　\\
　\\
余弦定理 \\
\begin{array}{l}
a^2 &= b^2 + c^2 -2bc \cos A \\
b^2 &= c^2 + a^2 -2ca \cos B \\
c^2 &= a^2 + b^2 -2ab \cos C \\
\end{array}

\\
\begin{array}{l}
\sin(x + y) &= \sin x \cos y + \cos x \sin y \\
\sin(x - y) &= \sin x \cos y - \cos x \sin y \\
\cos(x + y) &= \cos x \cos y - \sin x \sin y \\
\cos(x - y) &= \cos x \cos y + \sin x \sin y \\
\tan(x + y) &= \frac{\tan x + \tan y}{1 - \tan x \tan y} \\
\tan(x - y) &= \frac{\tan x - \tan y}{1 + \tan x \tan y} \\
\end{array}

\begin{array}{rl}
\sin x + \sin y &= 2 \sin(\frac{x + y}{2}) \cos(\frac{x - y}{2}) \\
\sin x - \sin y &= 2 \cos(\frac{x + y}{2}) \sin(\frac{x - y}{2}) \\
\cos x + \cos y &= 2 \cos(\frac{x + y}{2}) \cos(\frac{x - y}{2}) \\
\cos x - \cos y &= -2 \sin(\frac{x + y}{2}) \sin(\frac{x - y}{2}) \\
\end{array}

\begin{array}{rl}
a \sin x + b \cos x &= \sqrt{a^2 + b^2} \sin(x + \alpha) \\
\\
\alpha は以下を満たす角度 \\
\cos \alpha &= \frac{a}{\sqrt{a^2 + b^2}} \\
\sin \alpha &= \frac{b}{\sqrt{a^2 + b^2}} \\
\end{array}

\\
\begin{array}{l}
\sin 2x &= 2 \sin x \ cos x \\
\cos 2x &= \cos^2 x - \sin^2 x = 2 \cos^2 - 1 = 1- 2 \sin^2 x \\
\tan 2x &= \frac{2 \tan x}{1 - \tan^2 x} \\
\sin 3x &= -4 \sin^3 x + 3 \sin x \\
\cos 3x &= 4 \cos^3 x - 3 \cos x \\
\sin^2 x &= \frac{1 - \cos 2x}{2} \\
\cos^2 x &= \frac{1 + \cos 2x}{2} \\
\tan^2 x &= \frac{1 - \cos 2x}{1 + \cos 2x} \\
\end{array}