角が複雑な三角関数を含む方程式

数研出版｜数学Ⅱ[709] p.143 練習23

\({\small (1)}~\)\(\sin\left(\theta+\displaystyle\frac{\,\pi\,}{\,6\,}\right)=\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta+\displaystyle\frac{\,\pi\,}{\,6\,}\) の範囲は、

---

\(\begin{eqnarray}~~~\displaystyle\frac{\,\pi\,}{\,6\,}{\small ~≦~}\theta+\displaystyle\frac{\,\pi\,}{\,6\,}\lt 2\pi+\displaystyle\frac{\,\pi\,}{\,6\,}\end{eqnarray}\)

---

この範囲で、単位円と \(y=\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\) の交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(2\) 個分と \(4\) 個分であるので、

\(\begin{eqnarray}~~~\theta+\displaystyle\frac{\,\pi\,}{\,6\,}&=&\displaystyle\frac{\,\pi\,}{\,3\,}~,~\displaystyle\frac{\,2\,}{\,3\,}\pi\end{eqnarray}\)

---

よって、\(\theta+\displaystyle\frac{\,\pi\,}{\,6\,}=\displaystyle\frac{\,\pi\,}{\,3\,}\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,\pi\,}{\,3\,}-\displaystyle\frac{\,\pi\,}{\,6\,}
\\[3pt]~~~&=&\displaystyle\frac{\,2-1\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,\pi\,}{\,6\,}\end{eqnarray}\)

---

また、\(\theta+\displaystyle\frac{\,\pi\,}{\,6\,}=\displaystyle\frac{\,2\,}{\,3\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,2\,}{\,3\,}\pi-\displaystyle\frac{\,\pi\,}{\,6\,}
\\[3pt]~~~&=&\displaystyle\frac{\,4-1\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,\pi\,}{\,2\,}\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,\pi\,}{\,6\,}~,~\displaystyle\frac{\,\pi\,}{\,2\,}\)

 

\({\small (2)}~\)\(\cos\left(\theta-\displaystyle\frac{\,\pi\,}{\,3\,}\right)=-\displaystyle\frac{\,1\,}{\,\sqrt{2}\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta-\displaystyle\frac{\,\pi\,}{\,3\,}\) の範囲は、

---

\(\begin{eqnarray}~~~-\displaystyle\frac{\,\pi\,}{\,3\,}{\small ~≦~}\theta-\displaystyle\frac{\,\pi\,}{\,3\,}\lt 2\pi-\displaystyle\frac{\,\pi\,}{\,3\,}\end{eqnarray}\)

---

この範囲で、単位円と \(x=-\displaystyle\frac{\,1\,}{\,\sqrt{2}\,}\) の交点は、

---

---

\(1:1:\sqrt{2}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,4\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,4\,}\) が \(3\) 個分と \(5\) 個分であるので、

\(\begin{eqnarray}~~~\theta-\displaystyle\frac{\,\pi\,}{\,3\,}&=&\displaystyle\frac{\,3\,}{\,4\,}\pi~,~\displaystyle\frac{\,5\,}{\,4\,}\pi\end{eqnarray}\)

---

よって、\(\theta-\displaystyle\frac{\,\pi\,}{\,3\,}=\displaystyle\frac{\,3\,}{\,4\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,3\,}{\,4\,}\pi+\displaystyle\frac{\,\pi\,}{\,3\,}
\\[3pt]~~~&=&\displaystyle\frac{\,9+4\,}{\,12\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,13\,}{\,12\,}\pi\end{eqnarray}\)

---

また、\(\theta-\displaystyle\frac{\,\pi\,}{\,3\,}=\displaystyle\frac{\,5\,}{\,4\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,5\,}{\,4\,}\pi+\displaystyle\frac{\,\pi\,}{\,3\,}
\\[3pt]~~~&=&\displaystyle\frac{\,15+4\,}{\,12\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,19\,}{\,12\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,13\,}{\,12\,}\pi~,~\displaystyle\frac{\,19\,}{\,12\,}\pi\)

 

\({\small (3)}~\)\(\sin\left(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}\right)=\displaystyle\frac{\,1\,}{\,2\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}\) の範囲は、

---

\(\begin{eqnarray}~~~\displaystyle\frac{\,\pi\,}{\,4\,}{\small ~≦~}\theta+\displaystyle\frac{\,\pi\,}{\,4\,}\lt 2\pi+\displaystyle\frac{\,\pi\,}{\,4\,}\end{eqnarray}\)

---

この範囲で、単位円と \(y=\displaystyle\frac{\,1\,}{\,2\,}\) の交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(5\) 個分と \(13\) 個分であるので、

\(\begin{eqnarray}~~~\theta+\displaystyle\frac{\,\pi\,}{\,4\,}&=&\displaystyle\frac{\,5\,}{\,6\,}\pi~,~\displaystyle\frac{\,13\,}{\,6\,}\pi\end{eqnarray}\)

---

よって、\(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}=\displaystyle\frac{\,5\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,5\,}{\,6\,}\pi-\displaystyle\frac{\,\pi\,}{\,4\,}
\\[3pt]~~~&=&\displaystyle\frac{\,10-3\,}{\,12\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,7\,}{\,12\,}\pi\end{eqnarray}\)

---

また、\(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}=\displaystyle\frac{\,13\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,13\,}{\,6\,}\pi-\displaystyle\frac{\,\pi\,}{\,4\,}
\\[3pt]~~~&=&\displaystyle\frac{\,26-3\,}{\,12\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,23\,}{\,12\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,7\,}{\,12\,}\pi~,~\displaystyle\frac{\,23\,}{\,12\,}\pi\)

 

\({\small (4)}~\)\(\tan\left(\theta+\displaystyle\frac{\,\pi\,}{\,6\,}\right)=1\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta+\displaystyle\frac{\,\pi\,}{\,6\,}\) の範囲は、

---

\(\begin{eqnarray}~~~\displaystyle\frac{\,\pi\,}{\,6\,}{\small ~≦~}\theta+\displaystyle\frac{\,\pi\,}{\,6\,}\lt 2\pi+\displaystyle\frac{\,\pi\,}{\,6\,}\end{eqnarray}\)

---

この範囲で、直線 \(x=1\) 上の点 \((1~,~1)\) と原点を結ぶ直線と単位円との交点は、

---

---

\(1:1:\sqrt{2}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,4\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,4\,}\) が \(1\) 個分と \(5\) 個分であるので、

\(\begin{eqnarray}~~~\theta+\displaystyle\frac{\,\pi\,}{\,6\,}&=&\displaystyle\frac{\,\pi\,}{\,4\,}~,~\displaystyle\frac{\,5\,}{\,4\,}\pi\end{eqnarray}\)

---

よって、\(\theta+\displaystyle\frac{\,\pi\,}{\,6\,}=\displaystyle\frac{\,\pi\,}{\,4\,}\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,\pi\,}{\,4\,}-\displaystyle\frac{\,\pi\,}{\,6\,}
\\[3pt]~~~&=&\displaystyle\frac{\,3-2\,}{\,12\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,\pi\,}{\,12\,}\end{eqnarray}\)

---

また、\(\theta+\displaystyle\frac{\,\pi\,}{\,6\,}=\displaystyle\frac{\,5\,}{\,4\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,5\,}{\,4\,}\pi-\displaystyle\frac{\,\pi\,}{\,6\,}
\\[3pt]~~~&=&\displaystyle\frac{\,15-2\,}{\,12\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,13\,}{\,12\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,\pi\,}{\,12\,}~,~\displaystyle\frac{\,13\,}{\,12\,}\pi\)

数研出版｜数学Ⅱ[709] p.145 問題 4(4)

\(\sin\left(\theta-\displaystyle\frac{\,\pi\,}{\,3\,}\right)=-\displaystyle\frac{\,1\,}{\,2\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta-\displaystyle\frac{\,\pi\,}{\,3\,}\) の範囲は、

---

\(\begin{eqnarray}~~~-\displaystyle\frac{\,\pi\,}{\,3\,}{\small ~≦~}\theta-\displaystyle\frac{\,\pi\,}{\,3\,}\lt 2\pi-\displaystyle\frac{\,\pi\,}{\,3\,}\end{eqnarray}\)

---

この範囲で、単位円と \(y=-\displaystyle\frac{\,1\,}{\,2\,}\) の交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(-1\) 個分と \(7\) 個分であるので、

\(\begin{eqnarray}~~~\theta-\displaystyle\frac{\,\pi\,}{\,3\,}&=&-\displaystyle\frac{\,\pi\,}{\,6\,}~,~\displaystyle\frac{\,7\,}{\,6\,}\pi\end{eqnarray}\)

---

よって、\(\theta-\displaystyle\frac{\,\pi\,}{\,3\,}=-\displaystyle\frac{\,\pi\,}{\,6\,}\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&-\displaystyle\frac{\,\pi\,}{\,6\,}+\displaystyle\frac{\,\pi\,}{\,3\,}
\\[3pt]~~~&=&\displaystyle\frac{\,-1+2\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,\pi\,}{\,6\,}\end{eqnarray}\)

---

また、\(\theta-\displaystyle\frac{\,\pi\,}{\,3\,}=\displaystyle\frac{\,7\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,7\,}{\,6\,}\pi+\displaystyle\frac{\,\pi\,}{\,3\,}
\\[3pt]~~~&=&\displaystyle\frac{\,7+2\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,3\,}{\,2\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,\pi\,}{\,6\,}~,~\displaystyle\frac{\,3\,}{\,2\,}\pi\)

数研出版｜高等学校数学Ⅱ[710] p.131 練習20

\({\small (1)}~\)\(\sin\left(\theta-\displaystyle\frac{\,\pi\,}{\,6\,}\right)=-\displaystyle\frac{\,1\,}{\,\sqrt{2}\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta-\displaystyle\frac{\,\pi\,}{\,6\,}\) の範囲は、

---

\(\begin{eqnarray}~~~-\displaystyle\frac{\,\pi\,}{\,6\,}{\small ~≦~}\theta-\displaystyle\frac{\,\pi\,}{\,6\,}\lt 2\pi-\displaystyle\frac{\,\pi\,}{\,6\,}\end{eqnarray}\)

---

この範囲で、単位円と \(y=-\displaystyle\frac{\,1\,}{\,\sqrt{2}\,}\) の交点は、

---

---

\(1:1:\sqrt{2}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,4\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,4\,}\) が \(5\) 個分と \(7\) 個分であるので、

\(\begin{eqnarray}~~~\theta-\displaystyle\frac{\,\pi\,}{\,6\,}&=&\displaystyle\frac{\,5\,}{\,4\,}\pi~,~\displaystyle\frac{\,7\,}{\,4\,}\pi\end{eqnarray}\)

---

よって、\(\theta-\displaystyle\frac{\,\pi\,}{\,6\,}=\displaystyle\frac{\,5\,}{\,4\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,5\,}{\,4\,}\pi+\displaystyle\frac{\,\pi\,}{\,6\,}
\\[3pt]~~~&=&\displaystyle\frac{\,15+2\,}{\,12\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,17\,}{\,12\,}\pi\end{eqnarray}\)

---

また、\(\theta-\displaystyle\frac{\,\pi\,}{\,6\,}=\displaystyle\frac{\,7\,}{\,4\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,7\,}{\,4\,}\pi+\displaystyle\frac{\,\pi\,}{\,6\,}
\\[3pt]~~~&=&\displaystyle\frac{\,21+2\,}{\,12\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,23\,}{\,12\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,17\,}{\,12\,}\pi~,~\displaystyle\frac{\,23\,}{\,12\,}\pi\)

 

\({\small (2)}~\)\(\cos\left(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}\right)=\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}\) の範囲は、

---

\(\begin{eqnarray}~~~\displaystyle\frac{\,\pi\,}{\,4\,}{\small ~≦~}\theta+\displaystyle\frac{\,\pi\,}{\,4\,}\lt 2\pi+\displaystyle\frac{\,\pi\,}{\,4\,}\end{eqnarray}\)

---

この範囲で、単位円と \(x=\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\) の交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(11\) 個分と \(13\) 個分であるので、

\(\begin{eqnarray}~~~\theta+\displaystyle\frac{\,\pi\,}{\,4\,}&=&\displaystyle\frac{\,11\,}{\,6\,}\pi~,~\displaystyle\frac{\,13\,}{\,6\,}\pi\end{eqnarray}\)

---

よって、\(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}=\displaystyle\frac{\,11\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,11\,}{\,6\,}\pi-\displaystyle\frac{\,\pi\,}{\,4\,}
\\[3pt]~~~&=&\displaystyle\frac{\,22-3\,}{\,12\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,19\,}{\,12\,}\pi\end{eqnarray}\)

---

また、\(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}=\displaystyle\frac{\,13\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,13\,}{\,6\,}\pi-\displaystyle\frac{\,\pi\,}{\,4\,}
\\[3pt]~~~&=&\displaystyle\frac{\,26-3\,}{\,12\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,23\,}{\,12\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,19\,}{\,12\,}\pi~,~\displaystyle\frac{\,23\,}{\,12\,}\pi\)

数研出版｜高等学校数学Ⅱ[710] p.135 問題 3(5)

\(\cos\left(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}\right)=-\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}\) の範囲は、

---

\(\begin{eqnarray}~~~\displaystyle\frac{\,\pi\,}{\,3\,}{\small ~≦~}\theta+\displaystyle\frac{\,\pi\,}{\,3\,}\lt 2\pi+\displaystyle\frac{\,\pi\,}{\,3\,}\end{eqnarray}\)

---

この範囲で、単位円と \(x=-\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\) の交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(5\) 個分と \(7\) 個分であるので、

\(\begin{eqnarray}~~~\theta+\displaystyle\frac{\,\pi\,}{\,3\,}&=&\displaystyle\frac{\,5\,}{\,6\,}\pi~,~\displaystyle\frac{\,7\,}{\,6\,}\pi\end{eqnarray}\)

---

よって、\(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}=\displaystyle\frac{\,5\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,5\,}{\,6\,}\pi-\displaystyle\frac{\,\pi\,}{\,3\,}
\\[3pt]~~~&=&\displaystyle\frac{\,5-2\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,\pi\,}{\,2\,}\end{eqnarray}\)

---

また、\(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}=\displaystyle\frac{\,7\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,7\,}{\,6\,}\pi-\displaystyle\frac{\,\pi\,}{\,3\,}
\\[3pt]~~~&=&\displaystyle\frac{\,7-2\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,5\,}{\,6\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,\pi\,}{\,2\,}~,~\displaystyle\frac{\,5\,}{\,6\,}\pi\)

数研出版｜新編数学Ⅱ[711] p.132 補充問題 2

\({\small (1)}~\)\(\sin\left(\theta-\displaystyle\frac{\,\pi\,}{\,6\,}\right)=\displaystyle\frac{\,1\,}{\,2\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta-\displaystyle\frac{\,\pi\,}{\,6\,}\) の範囲は、

---

\(\begin{eqnarray}~~~-\displaystyle\frac{\,\pi\,}{\,6\,}{\small ~≦~}\theta-\displaystyle\frac{\,\pi\,}{\,6\,}\lt 2\pi-\displaystyle\frac{\,\pi\,}{\,6\,}\end{eqnarray}\)

---

この範囲で、単位円と \(y=\displaystyle\frac{\,1\,}{\,2\,}\) の交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(1\) 個分と \(5\) 個分であるので、

\(\begin{eqnarray}~~~\theta-\displaystyle\frac{\,\pi\,}{\,6\,}&=&\displaystyle\frac{\,\pi\,}{\,6\,}~,~\displaystyle\frac{\,5\,}{\,6\,}\pi\end{eqnarray}\)

---

よって、\(\theta-\displaystyle\frac{\,\pi\,}{\,6\,}=\displaystyle\frac{\,\pi\,}{\,6\,}\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,\pi\,}{\,6\,}+\displaystyle\frac{\,\pi\,}{\,6\,}
\\[3pt]~~~&=&\displaystyle\frac{\,\pi\,}{\,3\,}\end{eqnarray}\)

---

また、\(\theta-\displaystyle\frac{\,\pi\,}{\,6\,}=\displaystyle\frac{\,5\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,5\,}{\,6\,}\pi+\displaystyle\frac{\,\pi\,}{\,6\,}
\\[3pt]~~~&=&\displaystyle\frac{\,5+1\,}{\,6\,}\pi
\\[3pt]~~~&=&\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,\pi\,}{\,3\,}~,~\pi\)

 
 

\({\small (2)}~\)\(\cos\left(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}\right)=-\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}\) の範囲は、

---

\(\begin{eqnarray}~~~\displaystyle\frac{\,\pi\,}{\,3\,}{\small ~≦~}\theta+\displaystyle\frac{\,\pi\,}{\,3\,}\lt 2\pi+\displaystyle\frac{\,\pi\,}{\,3\,}\end{eqnarray}\)

---

この範囲で、単位円と \(x=-\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\) の交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(5\) 個分と \(7\) 個分であるので、

\(\begin{eqnarray}~~~\theta+\displaystyle\frac{\,\pi\,}{\,3\,}&=&\displaystyle\frac{\,5\,}{\,6\,}\pi~,~\displaystyle\frac{\,7\,}{\,6\,}\pi\end{eqnarray}\)

---

よって、\(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}=\displaystyle\frac{\,5\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,5\,}{\,6\,}\pi-\displaystyle\frac{\,\pi\,}{\,3\,}
\\[3pt]~~~&=&\displaystyle\frac{\,5-2\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,\pi\,}{\,2\,}\end{eqnarray}\)

---

また、\(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}=\displaystyle\frac{\,7\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,7\,}{\,6\,}\pi-\displaystyle\frac{\,\pi\,}{\,3\,}
\\[3pt]~~~&=&\displaystyle\frac{\,7-2\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,5\,}{\,6\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,\pi\,}{\,2\,}~,~\displaystyle\frac{\,5\,}{\,6\,}\pi\)

 
 

\({\small (3)}~\)\(\sin\left(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}\right)=\displaystyle\frac{\,1\,}{\,2\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}\) の範囲は、

---

\(\begin{eqnarray}~~~\displaystyle\frac{\,\pi\,}{\,3\,}{\small ~≦~}\theta+\displaystyle\frac{\,\pi\,}{\,3\,}\lt 2\pi+\displaystyle\frac{\,\pi\,}{\,3\,}\end{eqnarray}\)

---

この範囲で、単位円と \(y=\displaystyle\frac{\,1\,}{\,2\,}\) の交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(5\) 個分と \(13\) 個分であるので、

\(\begin{eqnarray}~~~\theta+\displaystyle\frac{\,\pi\,}{\,3\,}&=&\displaystyle\frac{\,5\,}{\,6\,}\pi~,~\displaystyle\frac{\,13\,}{\,6\,}\pi\end{eqnarray}\)

---

よって、\(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}=\displaystyle\frac{\,5\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,5\,}{\,6\,}\pi-\displaystyle\frac{\,\pi\,}{\,3\,}
\\[3pt]~~~&=&\displaystyle\frac{\,5-2\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,\pi\,}{\,2\,}\end{eqnarray}\)

---

また、\(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}=\displaystyle\frac{\,13\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,13\,}{\,6\,}\pi-\displaystyle\frac{\,\pi\,}{\,3\,}
\\[3pt]~~~&=&\displaystyle\frac{\,13-2\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,11\,}{\,6\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,\pi\,}{\,2\,}~,~\displaystyle\frac{\,11\,}{\,6\,}\pi\)

東京書籍｜Advanced数学Ⅱ[701] p.131 問25

\({\small (1)}~\)\(\cos\left(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}\right)=\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}\) の範囲は、

---

\(\begin{eqnarray}~~~\displaystyle\frac{\,\pi\,}{\,3\,}{\small ~≦~}\theta+\displaystyle\frac{\,\pi\,}{\,3\,}\lt 2\pi+\displaystyle\frac{\,\pi\,}{\,3\,}\end{eqnarray}\)

---

この範囲で、単位円と \(x=\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\) の交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(11\) 個分と \(13\) 個分であるので、

\(\begin{eqnarray}~~~\theta+\displaystyle\frac{\,\pi\,}{\,3\,}&=&\displaystyle\frac{\,11\,}{\,6\,}\pi~,~\displaystyle\frac{\,13\,}{\,6\,}\pi\end{eqnarray}\)

---

よって、\(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}=\displaystyle\frac{\,11\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,11\,}{\,6\,}\pi-\displaystyle\frac{\,\pi\,}{\,3\,}
\\[3pt]~~~&=&\displaystyle\frac{\,11-2\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,3\,}{\,2\,}\pi\end{eqnarray}\)

---

また、\(\theta+\displaystyle\frac{\,\pi\,}{\,3\,}=\displaystyle\frac{\,13\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,13\,}{\,6\,}\pi-\displaystyle\frac{\,\pi\,}{\,3\,}
\\[3pt]~~~&=&\displaystyle\frac{\,13-2\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,11\,}{\,6\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,3\,}{\,2\,}\pi~,~\displaystyle\frac{\,11\,}{\,6\,}\pi\)

 
 

\({\small (2)}~\)\(\sin\left(\theta-\displaystyle\frac{\,3\,}{\,4\,}\pi\right)=-\displaystyle\frac{\,1\,}{\,2\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta-\displaystyle\frac{\,3\,}{\,4\,}\pi\) の範囲は、

---

\(\begin{eqnarray}~~~-\displaystyle\frac{\,3\,}{\,4\,}\pi{\small ~≦~}\theta-\displaystyle\frac{\,3\,}{\,4\,}\pi\lt 2\pi-\displaystyle\frac{\,3\,}{\,4\,}\pi\end{eqnarray}\)

---

この範囲で、単位円と \(y=-\displaystyle\frac{\,1\,}{\,2\,}\) の交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(-1\) 個分と \(7\) 個分であるので、

\(\begin{eqnarray}~~~\theta-\displaystyle\frac{\,3\,}{\,4\,}\pi&=&-\displaystyle\frac{\,\pi\,}{\,6\,}~,~\displaystyle\frac{\,7\,}{\,6\,}\pi\end{eqnarray}\)

---

よって、\(\theta-\displaystyle\frac{\,3\,}{\,4\,}\pi=-\displaystyle\frac{\,\pi\,}{\,6\,}\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&-\displaystyle\frac{\,\pi\,}{\,6\,}+\displaystyle\frac{\,3\,}{\,4\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,-2+9\,}{\,12\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,7\,}{\,12\,}\pi\end{eqnarray}\)

---

また、\(\theta-\displaystyle\frac{\,3\,}{\,4\,}\pi=\displaystyle\frac{\,7\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,7\,}{\,6\,}\pi+\displaystyle\frac{\,3\,}{\,4\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,14+9\,}{\,12\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,23\,}{\,12\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,7\,}{\,12\,}\pi~,~\displaystyle\frac{\,23\,}{\,12\,}\pi\)

東京書籍｜Advanced数学Ⅱ[701] p.135 問題 6(2)(3)

\({\small (1)}~\)\(2\sin 2\theta=\sqrt{3}\) について、

---

両辺を \(2\) で割ると、

---

\(\begin{eqnarray}~~~\sin 2\theta=\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\end{eqnarray}\)

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(2\theta\) の範囲は、

---

\(\begin{eqnarray}~~~0{\small ~≦~}2\theta\lt 4\pi\end{eqnarray}\)

---

この範囲で、単位円と \(y=\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\) の交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(2\) 個分、\(4\) 個分、\(14\) 個分、\(16\) 個分であるので、

\(\begin{eqnarray}~~~2\theta&=&\displaystyle\frac{\,\pi\,}{\,3\,}~,~\displaystyle\frac{\,2\,}{\,3\,}\pi~,~\displaystyle\frac{\,7\,}{\,3\,}\pi~,~\displaystyle\frac{\,8\,}{\,3\,}\pi\end{eqnarray}\)

---

それぞれの値の両辺を \(2\) で割ると、

---

※ 数式は横にスクロールできます。

---

したがって、
　\(\theta=\displaystyle\frac{\,\pi\,}{\,6\,}~,~\displaystyle\frac{\,\pi\,}{\,3\,}~,~\displaystyle\frac{\,7\,}{\,6\,}\pi~,~\displaystyle\frac{\,4\,}{\,3\,}\pi\)

 
 

\({\small (2)}~\)\(\tan\displaystyle\frac{\,\theta\,}{\,2\,}+1=0\) について、

---

\(\tan\displaystyle\frac{\,\theta\,}{\,2\,}\) について整理すると、

---

\(\begin{eqnarray}~~~\tan\displaystyle\frac{\,\theta\,}{\,2\,}=-1\end{eqnarray}\)

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\displaystyle\frac{\,\theta\,}{\,2\,}\) の範囲は、

---

\(\begin{eqnarray}~~~0{\small ~≦~}\displaystyle\frac{\,\theta\,}{\,2\,}\lt\pi\end{eqnarray}\)

---

この範囲で、直線 \(x=1\) 上の点 \((1~,~-1)\) と原点を結ぶ直線と単位円との交点は、

---

---

\(1:1:\sqrt{2}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,4\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,4\,}\) が \(3\) 個分であるので、

\(\begin{eqnarray}~~~\displaystyle\frac{\,\theta\,}{\,2\,}&=&\displaystyle\frac{\,3\,}{\,4\,}\pi\end{eqnarray}\)

---

両辺を \(2\) 倍すると、

---

\(\begin{eqnarray}~~~\displaystyle\frac{\,\theta\,}{\,2\,}{\,\small \times \,}2&=&\displaystyle\frac{\,3\,}{\,4\,}\pi{\,\small \times \,}2
\\[5pt]~~~\theta&=&\displaystyle\frac{\,3\,}{\,2\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,3\,}{\,2\,}\pi\)

東京書籍｜Advanced数学Ⅱ[701] p.154 練習問題A 4(1)

\(2\cos\left(\displaystyle\frac{\,3\,}{\,2\,}\pi-\theta\right)=-1\) について、

---

両辺を \(2\) で割ると、

---

\(\begin{eqnarray}~~~\cos\left(\displaystyle\frac{\,3\,}{\,2\,}\pi-\theta\right)=-\displaystyle\frac{\,1\,}{\,2\,}\end{eqnarray}\)

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\displaystyle\frac{\,3\,}{\,2\,}\pi-\theta\) の範囲は、

---

\(\begin{eqnarray}~~~-2\pi\lt -\theta{\small ~≦~}0\end{eqnarray}\)

---

各辺に \(\displaystyle\frac{\,3\,}{\,2\,}\pi\) を加えると、

---

\(\begin{eqnarray}~~~-2\pi+\displaystyle\frac{\,3\,}{\,2\,}\pi &\lt& \displaystyle\frac{\,3\,}{\,2\,}\pi-\theta{\small ~≦~}\displaystyle\frac{\,3\,}{\,2\,}\pi
\\[5pt]-\displaystyle\frac{\,\pi\,}{\,2\,}&\lt& \displaystyle\frac{\,3\,}{\,2\,}\pi-\theta{\small ~≦~}\displaystyle\frac{\,3\,}{\,2\,}\pi\end{eqnarray}\)

---

この範囲で、単位円と \(x=-\displaystyle\frac{\,1\,}{\,2\,}\) の交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(4\) 個分と \(8\) 個分であるので、

\(\begin{eqnarray}~~~\displaystyle\frac{\,3\,}{\,2\,}\pi-\theta&=&\displaystyle\frac{\,2\,}{\,3\,}\pi~,~\displaystyle\frac{\,4\,}{\,3\,}\pi\end{eqnarray}\)

---

よって、\(\displaystyle\frac{\,3\,}{\,2\,}\pi-\theta=\displaystyle\frac{\,2\,}{\,3\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,3\,}{\,2\,}\pi-\displaystyle\frac{\,2\,}{\,3\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,9-4\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,5\,}{\,6\,}\pi\end{eqnarray}\)

---

また、\(\displaystyle\frac{\,3\,}{\,2\,}\pi-\theta=\displaystyle\frac{\,4\,}{\,3\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,3\,}{\,2\,}\pi-\displaystyle\frac{\,4\,}{\,3\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,9-8\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,\pi\,}{\,6\,}\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,\pi\,}{\,6\,}~,~\displaystyle\frac{\,5\,}{\,6\,}\pi\)

東京書籍｜Standard数学Ⅱ[702] p.140 問23

\({\small (1)}~\)\(\cos\left(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}\right)=-\displaystyle\frac{\,1\,}{\,\sqrt{2}\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}\) の範囲は、

---

\(\begin{eqnarray}~~~\displaystyle\frac{\,\pi\,}{\,4\,}{\small ~≦~}\theta+\displaystyle\frac{\,\pi\,}{\,4\,}\lt 2\pi+\displaystyle\frac{\,\pi\,}{\,4\,}\end{eqnarray}\)

---

この範囲で、単位円と \(x=-\displaystyle\frac{\,1\,}{\,\sqrt{2}\,}\) の交点は、

---

---

\(1:1:\sqrt{2}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,4\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,4\,}\) が \(3\) 個分と \(5\) 個分であるので、

\(\begin{eqnarray}~~~\theta+\displaystyle\frac{\,\pi\,}{\,4\,}&=&\displaystyle\frac{\,3\,}{\,4\,}\pi~,~\displaystyle\frac{\,5\,}{\,4\,}\pi\end{eqnarray}\)

---

よって、\(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}=\displaystyle\frac{\,3\,}{\,4\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,3\,}{\,4\,}\pi-\displaystyle\frac{\,\pi\,}{\,4\,}
\\[3pt]~~~&=&\displaystyle\frac{\,3-1\,}{\,4\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,\pi\,}{\,2\,}\end{eqnarray}\)

---

また、\(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}=\displaystyle\frac{\,5\,}{\,4\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,5\,}{\,4\,}\pi-\displaystyle\frac{\,\pi\,}{\,4\,}
\\[3pt]~~~&=&\displaystyle\frac{\,5-1\,}{\,4\,}\pi
\\[3pt]~~~&=&\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,\pi\,}{\,2\,}~,~\pi\)

 
 

\({\small (2)}~\)\(\sin\left(\theta-\displaystyle\frac{\,\pi\,}{\,3\,}\right)=-\displaystyle\frac{\,1\,}{\,2\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta-\displaystyle\frac{\,\pi\,}{\,3\,}\) の範囲は、

---

\(\begin{eqnarray}~~~-\displaystyle\frac{\,\pi\,}{\,3\,}{\small ~≦~}\theta-\displaystyle\frac{\,\pi\,}{\,3\,}\lt 2\pi-\displaystyle\frac{\,\pi\,}{\,3\,}\end{eqnarray}\)

---

この範囲で、単位円と \(y=-\displaystyle\frac{\,1\,}{\,2\,}\) の交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(-1\) 個分と \(7\) 個分であるので、

\(\begin{eqnarray}~~~\theta-\displaystyle\frac{\,\pi\,}{\,3\,}&=&-\displaystyle\frac{\,\pi\,}{\,6\,}~,~\displaystyle\frac{\,7\,}{\,6\,}\pi\end{eqnarray}\)

---

よって、\(\theta-\displaystyle\frac{\,\pi\,}{\,3\,}=-\displaystyle\frac{\,\pi\,}{\,6\,}\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&-\displaystyle\frac{\,\pi\,}{\,6\,}+\displaystyle\frac{\,\pi\,}{\,3\,}
\\[3pt]~~~&=&\displaystyle\frac{\,-1+2\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,\pi\,}{\,6\,}\end{eqnarray}\)

---

また、\(\theta-\displaystyle\frac{\,\pi\,}{\,3\,}=\displaystyle\frac{\,7\,}{\,6\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,7\,}{\,6\,}\pi+\displaystyle\frac{\,\pi\,}{\,3\,}
\\[3pt]~~~&=&\displaystyle\frac{\,7+2\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,3\,}{\,2\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,\pi\,}{\,6\,}~,~\displaystyle\frac{\,3\,}{\,2\,}\pi\)

東京書籍｜Standard数学Ⅱ[702] p.144 Training 8

\({\small (1)}~\)\(\sin\left(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}\right)=\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}\) の範囲は、

---

\(\begin{eqnarray}~~~\displaystyle\frac{\,\pi\,}{\,4\,}{\small ~≦~}\theta+\displaystyle\frac{\,\pi\,}{\,4\,}\lt 2\pi+\displaystyle\frac{\,\pi\,}{\,4\,}\end{eqnarray}\)

---

この範囲で、単位円と \(y=\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\) の交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(2\) 個分と \(4\) 個分であるので、

\(\begin{eqnarray}~~~\theta+\displaystyle\frac{\,\pi\,}{\,4\,}&=&\displaystyle\frac{\,\pi\,}{\,3\,}~,~\displaystyle\frac{\,2\,}{\,3\,}\pi\end{eqnarray}\)

---

よって、\(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}=\displaystyle\frac{\,\pi\,}{\,3\,}\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,\pi\,}{\,3\,}-\displaystyle\frac{\,\pi\,}{\,4\,}
\\[3pt]~~~&=&\displaystyle\frac{\,4-3\,}{\,12\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,\pi\,}{\,12\,}\end{eqnarray}\)

---

また、\(\theta+\displaystyle\frac{\,\pi\,}{\,4\,}=\displaystyle\frac{\,2\,}{\,3\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,2\,}{\,3\,}\pi-\displaystyle\frac{\,\pi\,}{\,4\,}
\\[3pt]~~~&=&\displaystyle\frac{\,8-3\,}{\,12\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,5\,}{\,12\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,\pi\,}{\,12\,}~,~\displaystyle\frac{\,5\,}{\,12\,}\pi\)

 
 

\({\small (2)}~\)\(\tan\left(\theta-\displaystyle\frac{\,\pi\,}{\,6\,}\right)=\sqrt{3}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(\theta-\displaystyle\frac{\,\pi\,}{\,6\,}\) の範囲は、

---

\(\begin{eqnarray}~~~-\displaystyle\frac{\,\pi\,}{\,6\,}{\small ~≦~}\theta-\displaystyle\frac{\,\pi\,}{\,6\,}\lt 2\pi-\displaystyle\frac{\,\pi\,}{\,6\,}\end{eqnarray}\)

---

この範囲で、直線 \(x=1\) 上の点 \((1~,~\sqrt{3})\) と原点を結ぶ直線と単位円との交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(2\) 個分と \(8\) 個分であるので、

\(\begin{eqnarray}~~~\theta-\displaystyle\frac{\,\pi\,}{\,6\,}&=&\displaystyle\frac{\,\pi\,}{\,3\,}~,~\displaystyle\frac{\,4\,}{\,3\,}\pi\end{eqnarray}\)

---

よって、\(\theta-\displaystyle\frac{\,\pi\,}{\,6\,}=\displaystyle\frac{\,\pi\,}{\,3\,}\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,\pi\,}{\,3\,}+\displaystyle\frac{\,\pi\,}{\,6\,}
\\[3pt]~~~&=&\displaystyle\frac{\,2+1\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,\pi\,}{\,2\,}\end{eqnarray}\)

---

また、\(\theta-\displaystyle\frac{\,\pi\,}{\,6\,}=\displaystyle\frac{\,4\,}{\,3\,}\pi\) のとき、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,4\,}{\,3\,}\pi+\displaystyle\frac{\,\pi\,}{\,6\,}
\\[3pt]~~~&=&\displaystyle\frac{\,8+1\,}{\,6\,}\pi
\\[3pt]~~~&=&\displaystyle\frac{\,3\,}{\,2\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,\pi\,}{\,2\,}~,~\displaystyle\frac{\,3\,}{\,2\,}\pi\)

東京書籍｜Standard数学Ⅱ[702] p.156 Level Up 5(1)

\(\sin\left(2\theta-\displaystyle\frac{\,\pi\,}{\,3\,}\right)=\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\) について、

---

\(0{\small ~≦~}\theta\lt 2\pi\) より、\(2\theta-\displaystyle\frac{\,\pi\,}{\,3\,}\) の範囲は、

---

\(\begin{eqnarray}~~~0{\small ~≦~}2\theta\lt 4\pi\end{eqnarray}\)

---

各辺から \(\displaystyle\frac{\,\pi\,}{\,3\,}\) を引くと、

---

\(\begin{eqnarray}~~~-\displaystyle\frac{\,\pi\,}{\,3\,}&{\small ~≦~}&2\theta-\displaystyle\frac{\,\pi\,}{\,3\,}\lt 4\pi-\displaystyle\frac{\,\pi\,}{\,3\,}\\[5pt]~~~-\displaystyle\frac{\,\pi\,}{\,3\,}&{\small ~≦~}&2\theta-\displaystyle\frac{\,\pi\,}{\,3\,}\lt \displaystyle\frac{\,11\,}{\,3\,}\pi\end{eqnarray}\)

---

この範囲で、単位円と \(y=\displaystyle\frac{\,\sqrt{3}\,}{\,2\,}\) の交点は、

---

---

\(1:2:\sqrt{3}\) の直角三角形の基本角 \(\displaystyle\frac{\,\pi\,}{\,6\,}\) より、\(\displaystyle\frac{\,\pi\,}{\,6\,}\) が \(2\) 個分、\(4\) 個分、\(14\) 個分、\(16\) 個分であるので、

\(\begin{eqnarray}~~~2\theta-\displaystyle\frac{\,\pi\,}{\,3\,}&=&\displaystyle\frac{\,\pi\,}{\,3\,}~,~\displaystyle\frac{\,2\,}{\,3\,}\pi~,~\displaystyle\frac{\,7\,}{\,3\,}\pi~,~\displaystyle\frac{\,8\,}{\,3\,}\pi\end{eqnarray}\)

---

それぞれに \(\displaystyle\frac{\,\pi\,}{\,3\,}\) を加えると、

---

※ 数式は横にスクロールできます。

---

それぞれ \(2\) で割ると、

---

\(\begin{eqnarray}~~~\theta&=&\displaystyle\frac{\,2\,}{\,3\,}\pi{\, \small \div \,}2~,~\pi{\, \small \div \,}2~,~\displaystyle\frac{\,8\,}{\,3\,}\pi{\, \small \div \,}2~,~3\pi{\, \small \div \,}2
\\[5pt]~~~\theta&=&\displaystyle\frac{\,\pi\,}{\,3\,}~,~\displaystyle\frac{\,\pi\,}{\,2\,}~,~\displaystyle\frac{\,4\,}{\,3\,}\pi~,~\displaystyle\frac{\,3\,}{\,2\,}\pi\end{eqnarray}\)

---

したがって、\(\theta=\displaystyle\frac{\,\pi\,}{\,3\,}~,~\displaystyle\frac{\,\pi\,}{\,2\,}~,~\displaystyle\frac{\,4\,}{\,3\,}\pi~,~\displaystyle\frac{\,3\,}{\,2\,}\pi\)