式展開: 9月 2024

問い(

\tan (θ) =

)

\overset{―}{A C}

に垂直な補助線

\overset{―}{B D}

を引く

\begin{array}{rcl} ∠ C & = & π - (∠ A + ∠ B) \\ = & π - {θ + (\frac{π}{2} + θ)} \\ = & \frac{π}{2} - 2 θ \end{array}

\begin{array}{rcl} \overset{―}{B D} & = & 2 \sin (\frac{π}{2} - 2 θ) \\ \overset{―}{C D} & = & 2 \cos (\frac{π}{2} - 2 θ) \\ \overset{―}{A D} & = & \overset{―}{A C} - \overset{―}{C D} \\ = & 5 - 2 \cos (\frac{π}{2} - 2 θ) \end{array}

\begin{array}{rcl} \tan (θ) & = & \frac{\overset{―}{B D}}{\overset{―}{A D}} \\ = & \frac{2 \sin (\frac{π}{2} - 2 θ)}{5 - 2 \cos (\frac{π}{2} - 2 θ)} \\ = & \frac{2 \cos (2 θ)}{5 - 2 \sin (2 θ)} \dots \cos (\frac{π}{2} - x) = \sin (x), \sin (\frac{π}{2} - x) = \cos (x) \\ = & \frac{2}{5} \dots \tan (θ) = \frac{B \cos (2 θ)}{A - B \sin (2 θ)} な ら ば \tan (θ) = \frac{B}{A} (下 記) \end{array}

\begin{array}{rcl} \tan (θ) & = & \frac{B \cos (2 θ)}{A - B \sin (2 θ)} \\ \frac{1}{\tan (θ)} & = & \frac{A - B \sin (2 θ)}{B \cos (2 θ)} \\ = & \frac{1}{\cos (2 θ)} {\frac{A}{B} - \frac{B}{B} \sin (2 θ)} \\ \frac{\cos (2 θ)}{\tan (θ)} & = & \frac{A}{B} - \sin (2 θ) \\ \frac{\cos (2 θ)}{\tan (θ)} + \sin (2 θ) & = & \frac{A}{B} \\ \frac{\cos (2 θ) + \tan (θ) \sin (2 θ)}{\tan (θ)} & = & \frac{A}{B} \\ \frac{\cos^{2} (θ) - \sin^{2} (θ) + 2 \tan (θ) \sin (θ) \cos (θ)}{\tan (θ)} & = & \frac{A}{B} \dots \cos (2 θ) = \cos^{2} (θ) - \sin^{2} (θ), \sin (2 θ) = 2 \sin (θ) \cos (θ) \\ \frac{\cos^{2} (θ) - \sin^{2} (θ) + 2 \frac{\sin (θ)}{\cos (θ)} \sin (θ) \cos (θ)}{\tan (θ)} & = & \frac{A}{B} \dots \tan (θ) = \frac{\sin (θ)}{\cos (θ)} \\ \frac{\cos^{2} (θ) - \sin^{2} (θ) + 2 \sin^{2} (θ)}{\tan (θ)} & = & \frac{A}{B} \\ \frac{\cos^{2} (θ) + \sin^{2} (θ)}{\tan (θ)} & = & \frac{A}{B} \\ \frac{1}{\tan (θ)} & = & \frac{A}{B} \dots \cos^{2} (θ) + \sin^{2} (θ) = 1 \\ \tan (θ) & = & \frac{B}{A} \end{array}

式展開