三角関数を含む不等式
課題２
\(0 \leqq x < 2\pi\) のとき，次の不等式を解きましょう。
1. \(\displaystyle \sin x \geqq \frac{\sqrt{3}}{2}\) 解答 隠す
\[\frac{\pi}{3} \leqq x \leqq \frac{2\pi}{3}\]
\(x\)
\(y\)
\(\displaystyle y = \frac{\sqrt{3}}{2}\)
2. \(4\sin^2 x - 3 < 0\) 解答 隠す
\[\begin{eqnarray}
&& \left(2\sin x + \sqrt{3}\right)\left(2\sin x - \sqrt{3}\right) < 0 \\[2px]
&& ∴\quad -\frac{\sqrt{3}}{2} < \sin x < \frac{\sqrt{3}}{2} \\[2px]
&& ∴\quad 0 \leqq x < \frac{\pi}{3},\ \frac{2\pi}{3} < x < \frac{4\pi}{3},\ \frac{5\pi}{3} < x < 2\pi
\end{eqnarray}\]
\(x\)
\(y\)
\(\displaystyle y = \frac{\sqrt{3}}{2}\)
\(\displaystyle y = -\frac{\sqrt{3}}{2}\)
3. \(4\cos^2 x + 2(\sqrt{3} - 1)\cos x - \sqrt{3} \geqq 0\) 解答 隠す
\[\begin{eqnarray}
&& (2\cos x + \sqrt{3})(2\cos x - 1) \geqq 0 \\[2px]
&& ∴\quad \cos x \leqq -\frac{\sqrt{3}}{2},\ \frac{1}{2} \leqq \cos x \\[2px]
&& ∴\quad 0 \leqq x \leqq \frac{\pi}{3},\ \frac{5\pi}{6} \leqq x \leqq \frac{7\pi}{6},\ \frac{5\pi}{3} \leqq x < 2\pi
\end{eqnarray}\]
\(x\)
\(y\)
\(\displaystyle x = \frac{1}{2}\)
\(\displaystyle x = -\frac{\sqrt{3}}{2}\)
三角関数を含む関数の最大・最小
課題３
\(0 \leqq x < 2\pi\) のとき，次の関数の最大値と最小値を求めましょう。
1. \(y = 1 + \cos x - \sin^2 x\) 解答 隠す
\(y = \cos^2 x + \cos x\) と変形できるから \(\cos x = t\) とおくと，\(-1 \leqq t \leqq 1\) であり
\[y = t^2 + t = \left(t + \frac{1}{2}\right)^2 - \frac{1}{4}\]
\(t = 1\) すなわち \(\cos x = 1\) のとき \(x = 0\)
\(\displaystyle t = -\frac{1}{2}\) すなわち \(\displaystyle \cos x = -\frac{1}{2}\) のとき \(\displaystyle x = \frac{2\pi}{3},\ \frac{4\pi}{3}\)
\(t\)
\(y\)
\[\left\{\begin{array}{l}
\mbox{Maximum} = 2 \\
\hspace{2em}(x = 0)\\[8px]
\displaystyle \mbox{Mimimum} = -\frac{1}{4} \\
\displaystyle \hspace{2em} \left(x = \frac{2\pi}{3},\ \frac{4\pi}{3}\right)
\end{array}\right.\]
2. \(y = -\cos^2 x - \displaystyle \frac{1}{2}\sin x\) 解答 隠す
\(\displaystyle y = \sin^2 x - \frac{1}{2}\sin x - 1\) と変形できるから \(\sin x = t\) とおくと，\(-1 \leqq t \leqq 1\) であり
\[y = t^2 - \frac{1}{2}t - 1 = \left(t - \frac{1}{4}\right)^2 - \frac{17}{16}\]
\(t = -1\) すなわち \(\sin x = -1\) のとき \(\displaystyle x = \frac{3\pi}{2}\)
\(\displaystyle t = \frac{1}{4}\) すなわち \(\displaystyle \sin x = \frac{1}{4}\) のとき \(\displaystyle \sin\phi = \frac{1}{4}\ \left(0 < \phi < \frac{\pi}{2}\right)\) とすると \(x = \phi,\ \pi - \phi\)
\[∴\quad\left\{\begin{array}{ll}
\displaystyle \mbox{Maximum} = \frac{1}{2} & \displaystyle \left(x = \frac{3\pi}{2}\right) \\
\displaystyle \mbox{Minimum} = -\frac{17}{16} & \displaystyle \left(x = \phi,\ \pi - \phi\right)
\end{array}\right.\\
\hspace{2em}\left(\sin\phi = \frac{1}{4},\ 0 < \phi < \frac{\pi}{2}\right)\]
3. \(y = \cos 2x + \sin x\) 解答 隠す
\(\displaystyle y = -2\sin^2 x + \sin x + 1\) と変形できるから \(\sin x = t\) とおくと，\(-1 \leqq t \leqq 1\) であり
\[y = -2t^2 + t + 1 = -2\left(t - \frac{1}{4}\right)^2 + \frac{9}{8}\]
\(t = -1\) すなわち \(\sin x = -1\) のとき \(\displaystyle x = \frac{3\pi}{2}\)
\(\displaystyle t = \frac{1}{4}\) すなわち \(\displaystyle \sin x = \frac{1}{4}\) のとき \(\displaystyle \sin\phi = \frac{1}{4}\ \left(0 < \phi < \frac{\pi}{2}\right)\) とすると \(x = \phi,\ \pi - \phi\)
\[∴\quad\left\{\begin{array}{ll}
\displaystyle \mbox{Maximum} = \frac{9}{8} & \displaystyle \left(x = \phi,\ \pi - \phi\right) \\
\displaystyle \mbox{Minimum} = -2 & \displaystyle \left(x = \frac{3\pi}{2}\right)
\end{array}\right.\\
\hspace{2em}\left(\sin\phi = \frac{1}{4},\ 0 < \phi < \frac{\pi}{2}\right)\]
4. \(y = 2\cos^2 x + 2\sin x\cos x - \sin^2 x\) 解答 隠す
\[\begin{eqnarray}
y &=& 1 + \cos 2x + \sin 2x - \frac{1 - \cos 2x}{2} \\[2px]
&=& \sin 2x + \frac{3}{2}\,\cos 2x + \frac{1}{2} \\[2px]
&=& \sqrt{1^2 + \left(\frac{3}{2}\right)^2}\sin(2x + \phi) + \frac{1}{2} \\[2px]
&=& \frac{\sqrt{13}}{2}\sin(2x + \phi) + \frac{1}{2} \\
&& \left(\cos\phi = \frac{2}{\sqrt{13}},\ \sin\phi = \frac{3}{\sqrt{13}}\right)
\end{eqnarray}\]
\(y\) が最大になるのは，\(\sin(2x + \phi) = 1\) となるときであり，\(\phi \leqq 2x + \phi < 4\pi + \phi\) なので，\(\displaystyle 2x + \phi = \frac{\pi}{2},\ \frac{5\pi}{2}\) すなわち \(\displaystyle x = \frac{\pi}{4} - \frac{\phi}{2},\ \frac{5\pi}{4} - \frac{\phi}{2}\)
\(y\) が最小になるのは，\(\sin(2x + \phi) = -1\) となるときであり，上記と同様に \(\displaystyle 2x + \phi = \frac{3\pi}{2},\ \frac{7\pi}{2}\) すなわち \(\displaystyle x = \frac{3\pi}{4} - \frac{\phi}{2},\ \frac{7\pi}{4} - \frac{\phi}{2}\)
\[∴\quad \left\{\begin{array}{ll}
\displaystyle \mbox{Maximum} = \frac{1}{2} + \frac{\sqrt{13}}{2} & \displaystyle \left(x = \frac{\pi}{4} - \frac{\phi}{2},\ \frac{5\pi}{4} - \frac{\phi}{2}\right) \\[2px]
\displaystyle \mbox{Minimum} = \frac{1}{2} - \frac{\sqrt{13}}{2} & \displaystyle \left(x = \frac{3\pi}{4} - \frac{\phi}{2},\ \frac{7\pi}{4} - \frac{\phi}{2}\right)
\end{array}\right.\\ \hspace{2em}\left(\cos\phi = \frac{2}{\sqrt{13}},\ \sin\phi = \frac{3}{\sqrt{13}}\right)\]