151116 初版 151116 更新

\(\sin\left(\theta+\dfrac{\pi}{4}\right)=\dfrac{1}{2}\) を満たす θ
一般には,
\(\theta+\dfrac{\pi}{4} = \dfrac{\pi}{6}+2n\pi\),  \(\dfrac{5}{6}\pi+2n\pi\)   (n は整数)
すなわち,
\(\theta = \dfrac{-1+24n}{12}\pi\),  \(\dfrac{7+24n}{12}\pi\)   (n は整数)
0 ≦ θ < 2π では,
\(\theta = \dfrac{7}{12}\pi\),  \(\dfrac{23}{12}\pi\)  

\(\sin\left(\theta+\dfrac{\pi}{4}\right) >\dfrac{1}{2}\) を満たす θ
一般には,
\(\dfrac{\pi}{6}+2n\pi < \theta+\dfrac{\pi}{4}\) \(< \dfrac{5}{6}\pi+2n\pi\)   (n は整数)
すなわち,
\(\dfrac{-1+24n}{12}\pi < \theta\) \(< \dfrac{7+24n}{12}\pi\)   (n は整数)
0 ≦ θ < 2π では,
\(0\leqq \theta < \dfrac{7}{12}\pi\),  \(\dfrac{23}{12}\pi < \theta < 2\pi\)

\(\sin\left(\theta+\dfrac{\pi}{4}\right) <\dfrac{1}{2}\) を満たす θ
一般には,
\(\dfrac{5}{6}\pi+2n\pi < \theta+\dfrac{\pi}{4}\) \(< \dfrac{13}{6}\pi+2n\pi\)   (n は整数)
すなわち,
\(\dfrac{7+24n}{12}\pi < \theta\) \(< \dfrac{23+24n}{12}\pi\)   (n は整数)
0 ≦ θ < 2π では,
\(\dfrac{7}{12}\pi < \theta\) \(< \dfrac{23}{12}\pi\)  

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