統計学にあまりでない数学

三角関数

\sin^{2} θ + \cos^{2} θ = 1

$\begin{equation} \sin^2\theta + \cos^2\theta = 1 \end{equation}$

\tan θ = \frac{\sin θ}{\cos θ}

$\begin{equation} \tan\theta = \frac{\sin\theta}{\cos\theta} \end{equation}$

1 + \tan^{2} θ = \frac{1}{\cos^{2} θ}

$\begin{equation} 1 + \tan^2\theta = \frac{1}{\cos^2\theta} \end{equation}$

\sin (α \pm β) = \sin α \cos β \pm \cos α \cos β

$\begin{equation} \sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\cos\beta \end{equation}$

\cos (α \pm β) = \cos α \cos β \mp \sin α \sin β

$\begin{equation} \cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta \end{equation}$

\tan (α \pm β) = \frac{\tan α \pm \tan β}{1 \mp \tan α \tan β}

$\begin{equation} \tan(\alpha \pm \beta) = \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta} \end{equation}$

\sin^{2} \frac{θ}{2} = \frac{1 - \cos θ}{2}

$\begin{equation} \sin^{2}\frac{\theta}{2} = \frac{1-\cos\theta}{2} \end{equation}$

\cos^{2} \frac{θ}{2} = \frac{1 + \cos θ}{2}

$\begin{equation} \cos^{2}\frac{\theta}{2} = \frac{1+\cos\theta}{2} \end{equation}$

\sin 2 θ = 2 \sin θ \cos θ

$\begin{equation} \sin 2\theta = 2\sin\theta\cos\theta \end{equation}$

\begin{aligned} \cos 2 θ & = \cos^{2} θ - \sin^{2} θ \\ = 2 \cos^{2} θ - 1 \\ = 1 - 2 \sin^{2} θ \end{aligned}

$\begin{align} \cos 2\theta &= \cos^{2}\theta - \sin^{2}\theta \\ &= 2\cos^{2}\theta - 1 \\ &= 1 - 2\sin^{2}\theta \end{align}$

\sin 3 θ = 3 \sin θ - 4 \sin^{3} θ

$\begin{equation} \sin 3\theta = 3\sin\theta - 4\sin^{3}\theta \end{equation}$

\cos 3 θ = 4 \cos^{3} θ - 3 \cos θ

$\begin{equation} \cos 3\theta = 4\cos^{3}\theta - 3\cos\theta \end{equation}$

\cos α \cos β = \frac{\cos (α - β) + \cos (α + β)}{2}

$\begin{equation} \cos\alpha \cos\beta = {\cos(\alpha - \beta) + \cos(\alpha + \beta) \over 2} \end{equation}$

\sin α \sin β = \frac{\cos (α - β) - \cos (α + β)}{2}

$\begin{equation} \sin\alpha \sin\beta = {\cos(\alpha - \beta) - \cos(\alpha + \beta) \over 2} \end{equation}$

\sin α \cos β = \frac{\sin (α + β) + \sin (α - β)}{2}

$\begin{equation} \sin\alpha \cos\beta = {\sin(\alpha + \beta) + \sin(\alpha - \beta) \over 2} \end{equation}$

\cos α \sin β = \frac{\sin (α + β) - \sin (α - β)}{2}

$\begin{equation} \cos\alpha \sin\beta = {\sin(\alpha + \beta) - \sin(\alpha - \beta) \over 2} \end{equation}$

\cos α + \cos β = 2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2})

$\begin{equation} \cos\alpha + \cos\beta = 2 \cos\left( \frac{\alpha + \beta}{2} \right) \cos\left( \frac{\alpha - \beta}{2} \right) \end{equation}$

\cos α - \cos β = - 2 \sin (\frac{α + β}{2}) \sin (\frac{α - β}{2})

$\begin{equation} \cos\alpha - \cos\beta = -2\sin\left( \frac{\alpha + \beta}{2} \right) \sin\left(\frac{\alpha - \beta}{2}\right) \end{equation}$

\sin α + \sin β = 2 \sin (\frac{α + β}{2}) \cos (\frac{α - β}{2})

$\begin{equation} \sin\alpha + \sin\beta = 2 \sin\left( \frac{\alpha + \beta}{2} \right) \cos\left( \frac{\alpha - \beta}{2} \right) \end{equation}$

\sin α - \sin β = 2 \sin (\frac{α - β}{2}) \cos (\frac{α + β}{2})

$\begin{equation} \sin\alpha - \sin\beta = 2 \sin\left( \frac{\alpha - \beta}{2} \right) \cos\left( \frac{\alpha + \beta}{2} \right) \end{equation}$

a \sin θ + b \cos θ = \sqrt{a^{2} + b^{2}} \sin (θ + α)

$\begin{equation} a\sin\theta + b\cos\theta = \sqrt{a^{2} + b^{2}}\sin(\theta + \alpha) \end{equation}$

\cos α = \frac{a}{\sqrt{a^{2} + b^{2}}}

$\begin{equation} \cos\alpha = \frac{a}{\sqrt{a^{2} + b^{2}}} \end{equation}$

\sin α = \frac{b}{\sqrt{a^{2} + b^{2}}}

$\begin{equation} \sin\alpha = \frac{b}{\sqrt{a^{2} + b^{2}}} \end{equation}$