三角関数と双曲線関数の2倍角と3倍角公式
三角関数の2倍角公式
(1)
\[ \sin2x=2\sin x\cos x \](2)
\begin{align*} \cos2x & =2\cos^{2}x-1\\ & =1-2\sin^{2}x\\ & =\cos^{2}x-\sin^{2}x \end{align*}(3)
\[ \tan2x=\frac{2\tan x}{1-\tan^{2}x} \](1)
\begin{align*} \sin2x & =\sin(x+x)\\ & =\sin x\cos x+\cos x\sin x\\ & =2\sin x\cos x \end{align*}(2)
\begin{align*} \cos2x & =\cos(x+x)\\ & =\cos x\cos x-\sin x\sin x\\ & =\cos^{2}x-\sin^{2}x\tag{*}\\ & =2\cos^{2}x-1\tag{*}\\ & =1-2\sin^{2}x\tag{*} \end{align*}(3)
\begin{align*} \tan2x & =\tan(x+x)\\ & =\frac{\tan x+\tan x}{1-\tan x\tan x}\\ & =\frac{2\tan x}{1-\tan^{2}x} \end{align*}双曲線関数の2倍角公式
(1)
\[ \sinh2x=2\sinh x\cosh x \](2)
\begin{align*} \cosh2x & =2\cosh^{2}x-1\\ & =1+2\sinh^{2}x\\ & =\cosh^{2}x+\sinh^{2}x \end{align*}(3)
\[ \tanh2x=\frac{2\tanh x}{1+\tanh^{2}x} \](1)
\begin{align*} \sinh2x & =-i\sin(2ix)\\ & =-2i\sin(ix)\cos(ix)\\ & =2\sinh x\cosh x \end{align*}(2)
\begin{align*} \cosh2x & =\cos(2ix)\\ & =2\cos^{2}(ix)-1\\ & =2\cosh^{2}x-1\tag{*}\\ & =1+2\sinh^{2}x\tag{*}\\ & =\cosh^{2}x+\sinh^{2}x\tag{*} \end{align*}(3)
\begin{align*} \tanh2x & =-i\tan(2ix)\\ & =-i\frac{2\tan(ix)}{1-\tan^{2}(ix)}\\ & =\frac{2\tanh x}{1+\tanh^{2}x} \end{align*}三角関数の3倍角公式
(1)
\[ \sin3x=3\sin x-4\sin^{3}x \](2)
\[ \cos3x=4\cos^{3}x-3\cos x \](3)
\[ \tan3x=\frac{3\tan x-\tan^{3}x}{1-3\tan^{2}x} \](1)
\begin{align*} \sin3x & =\sin(x+2x)\\ & =\sin x\cos2x+\cos x\sin2x\\ & =\sin x(1-2\sin^{2}x)+\cos x(2\sin x\cos x)\\ & =3\sin x-4\sin^{3}x \end{align*}(2)
\begin{align*} \cos3x & =\cos(x+2x)\\ & =\cos x\cos2x-\sin x\sin2x\\ & =\cos x(2\cos^{2}x-1)-\sin x(2\sin x\cos x)\\ & =4\cos^{3}x-3\cos x \end{align*}(3)
\begin{align*} \tan3x & =\frac{\sin3x}{\cos3x}\\ & =\frac{3\sin x-4\sin^{3}x}{4\cos^{3}x-3\cos x}\\ & =\frac{3\tan x\cos^{-2}x-4\tan^{3}x}{4-3\cos^{-2}x}\\ & =\frac{3\tan x(1+\tan^{2}x)-4\tan^{3}x}{4-3(1+\tan^{2}x)}\\ & =\frac{3\tan x-\tan^{3}x}{1-3\tan^{2}x} \end{align*}双曲線関数の3倍角公式
(1)
\[ \sinh3x=3\sinh x+4\sinh^{3}x \](2)
\[ \cosh3x=4\cosh^{3}x-3\cosh x \](3)
\[ \tanh3x=\frac{3\tanh x+\tanh^{3}x}{1+3\tanh^{2}x} \](1)
\begin{align*} \sinh3x & =-i\sin3ix\\ & =-i\left\{ 3\sin(ix)-4\sin^{3}(ix)\right\} \\ & =3\sinh x+4\sinh^{3}x \end{align*}(2)
\begin{align*} \cosh3x & =\cos(3ix)\\ & =4\cos^{3}(ix)-3\cos(ix)\\ & =4\cosh^{3}x--3\cosh x \end{align*}(3)
\begin{align*} \tanh3x & =-i\tan(3ix)\\ & =-i\frac{3\tan(ix)-\tan^{3}(ix)}{1-3\tan^{2}(ix)}\\ & =\frac{3\tanh x+\tanh^{3}x}{1+3\tanh^{2}x} \end{align*}ページ情報
タイトル | 三角関数と双曲線関数の2倍角と3倍角公式 |
URL | https://www.nomuramath.com/wj8itk5o/ |
SNSボタン |
逆正接関数・逆双曲線正接関数と多重対数関数の関係
\[
\Tan^{\bullet}z=\frac{i}{2}\left(-\Li_{1}\left(iz\right)+\Li_{1}\left(-iz\right)\right)
\]
1±itan(z)など
\[
1\pm i\tan z=\frac{1}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\left(e^{\pm2i\Re z}+e^{\mp2\Im z}\right)
\]
逆三角関数と逆双曲線関数の冪乗積分漸化式
\[
\int\sin^{\bullet,n}xdx=x\sin^{\bullet,n}x+n\sqrt{1-x^{2}}\sin^{\bullet,n-1}x-n(n-1)\int\sin^{\bullet,n-2}xdx
\]
三角関数(双曲線関数)の逆三角関数(逆双曲線関数)が恒等写像になる条件
\[
\sin^{\bullet}\sin z=?z
\]