高三のヨーコ参上まだ未婚 \cos3\theta$}{\textcolor{magenta}{ 高三}}\ \rubyb{$=$}{\textcolor{magenta}{の}}\ \rubyb{\textcolor{red}{4}}{\textcolor{magenta}{ヨ}}\rubyb{$\textcolor{red}{\cos^3\theta}$}{\textcolor{magenta}{ーコ参上}}\rubyb{$\textcolor{red}{-}$}{\textcolor{magenta}{まだ}}\rubyb{\textcolor{red}{3}}{\textcolor{magenta}{未}}\rubyb{$\textcolor{red}{\cos\theta}$}{\textcolor{magenta}{婚}}}$}} \\[.5zh] {\Large \textbf{\maru2\ \ $\bm{\rubyb{$\sin3\theta$}{\textcolor{magenta}{ }}\ \rubyb{$=$}{\textcolor{magenta}{ }}\textcolor{red}{\ -\ }\rubyb{\textcolor{red}{4}}{\textcolor{magenta}{ }}\rubyb{$\textcolor{red}{\sin^3\theta}$}{\textcolor{magenta}{ }}\rubyb{$\textcolor{red}{+}$}{\textcolor{magenta}{ }}\rubyb{\textcolor{red}{3}}{\textcolor{magenta}{ }}\rubyb{$\textcolor{red}{\sin\theta}$}{\textcolor{magenta}{ }}}$}} 先に,\ \cos3\theta\,をゴロ合わせで暗記しよう. \\[.2zh] \bm{\sin3\theta\,は,\ \cos\theta\ →\ \sin\theta\ として正負を逆にする}と覚えておけばよい. \\[.2zh] 以下のように加法定理から導けるが,\ 試験時間内でやる余裕はない. 3倍角の公式の導出}} \\[1zh] $\bm{\textcolor{red}{3\theta\ →\ 2\theta+\theta}}\ として,\ \bm{\textcolor{cyan}{加法定理}}と\bm{\textcolor{cyan}{2倍角の公式}}を利用する.$ \\\\ $\cos3\theta=\cos(\textcolor{red}{2\theta+\theta})=\textcolor{cyan}{\cos2\theta\cos\theta-\sin2\theta\sin\theta}$ {\small $[\textcolor{brown}{\,加法定理\,}]$} \\[.2zh] $\phantom{\cos3\theta}=(\textcolor{cyan}{2\cos^2\theta-1})\cos\theta-\textcolor{cyan}{2\sin\theta\cos\theta}\cdot\sin\theta$ \hspace{.4zw}{\small $[\textcolor{brown}{\,2倍角の公式\,}]$} \\[.2zh] $\phantom{\cos3\theta}=2\cos^3\theta-\cos\theta-2\sin^2\theta\cos\theta$ \\[.2zh] $\phantom{\cos3\theta}=2\cos^3\theta-\cos\theta-2(1-\cos^2\theta)\cos\theta$ \\[.2zh] $\phantom{\cos3\theta}=\bm{4\cos^3\theta-3\cos\theta}$ \\\\ $\sin3\theta=\sin(\textcolor{red}{2\theta+\theta})=\textcolor{cyan}{\sin2\theta\cos\theta+\cos2\theta\sin\theta}$ {\small $[\textcolor{brown}{\,加法定理\,}]$} \\[.2zh] $\phantom{\sin3\theta}=\textcolor{cyan}{2\sin\theta\cos\theta}\cdot\cos\theta+(\textcolor{cyan}{1-2\sin^2\theta})\sin\theta$ \hspace{.4zw}{\small $[\textcolor{brown}{\,2倍角の公式\,}]$} \\[.2zh] $\phantom{\sin3\theta}=2\sin\theta\cos^2\theta+\sin\theta-2\sin^3\theta$ \\[.2zh] $\phantom{\sin3\theta}=2\sin\theta(1-\sin^2\theta)+\sin\theta-2\sin^3\theta$ \\[.2zh] $\phantom{\sin3\theta}=\bm{-\,4\sin^3\theta+3\sin\theta}$ \cos3\theta\,は\,\cos\theta\,のみ,\ \sin3\theta\,は\,\sin\theta\,のみの式にすることを目指す. 受験数学ではまず見かけないが(暗記不必要),\ $\tan$の3倍角の公式も示しておく. \\[1zh] $\bm{\tan3\theta}=\tan(\textcolor{red}{2\theta+\theta})=\textcolor{cyan}{\bunsuu{\tan2\theta+\tan\theta}{1-\tan2\theta\tan\theta}}$ {\small $[\textcolor{brown}{\,加法定理\,}]$} \\[.5zh] $\phantom{\bm{\tan3\theta}}=\bunsuu{\bunsuu{2\tan\theta}{1-\tan^2\theta}+\tan\theta}{1-\bunsuu{2\tan\theta}{1-\tan^2\theta}\cdot\tan\theta}$ {\small $[\textcolor{brown}{\,もう一度加法定理\,}]$} \\[.5zh] $\phantom{\bm{\tan3\theta}}=\bunsuu{2\tan\theta+\tan\theta(1-\tan^2\theta)}{(1-\tan^2\theta)-2\tan^2\theta}$ {\small $[\textcolor{brown}{\,分母分子に1-\tan^2\theta\,を掛けた\,}]$} \\[.5zh] $\phantom{\bm{\tan3\theta}}=\bm{\bunsuu{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}}$
