加法定理を用いて「展開してみる」
\(\sin\left(\theta+\dfrac{\pi}{6}\right)=\sin\theta\cos\dfrac{\pi}{6}+\cos\theta\sin\dfrac{\pi}{6}\) \(=\dfrac{\sqrt{3}}{2}\sin\theta+\dfrac{1}{2}\cos\theta\)
\(\sin\left(\theta+\dfrac{\pi}{4}\right)=\sin\theta\cos\dfrac{\pi}{4}+\cos\theta\sin\dfrac{\pi}{4}\) \(=\dfrac{\sqrt{2}}{2}\sin\theta+\dfrac{\sqrt{2}}{2}\cos\theta\)
\(\sin\left(\theta+\dfrac{\pi}{3}\right)=\sin\theta\cos\dfrac{\pi}{3}+\cos\theta\sin\dfrac{\pi}{3}\) \(=\dfrac{1}{2}\sin\theta+\dfrac{\sqrt{3}}{2}\cos\theta\)
\(\sin\left(\theta+\dfrac{\pi}{2}\right)=\sin\theta\cos\dfrac{\pi}{2}+\cos\theta\sin\dfrac{\pi}{2}\) \(=\cos\theta\)

加法定理を用いて「展開してみる」
\(\sin\left(\theta-\dfrac{\pi}{6}\right)=\sin\theta\cos\dfrac{\pi}{6}-\cos\theta\sin\dfrac{\pi}{6}\) \(=\dfrac{\sqrt{3}}{2}\sin\theta-\dfrac{1}{2}\cos\theta\)
\(\sin\left(\theta-\dfrac{\pi}{4}\right)=\sin\theta\cos\dfrac{\pi}{4}-\cos\theta\sin\dfrac{\pi}{4}\) \(=\dfrac{\sqrt{2}}{2}\sin\theta-\dfrac{\sqrt{2}}{2}\cos\theta\)
\(\sin\left(\theta-\dfrac{\pi}{3}\right)=\sin\theta\cos\dfrac{\pi}{3}-\cos\theta\sin\dfrac{\pi}{3}\) \(=\dfrac{1}{2}\sin\theta-\dfrac{\sqrt{3}}{2}\cos\theta\)
\(\sin\left(\theta-\dfrac{\pi}{2}\right)=\sin\theta\cos\dfrac{\pi}{2}-\cos\theta\sin\dfrac{\pi}{2}\) \(=-\cos\theta\)

加法定理を用いて「展開してみる」
\(\cos\left(\theta-\dfrac{\pi}{6}\right)=\cos\theta\cos\dfrac{\pi}{6}+\sin\theta\sin\dfrac{\pi}{6}\) \(=\dfrac{1}{2}\sin\theta+\dfrac{\sqrt{3}}{2}\cos\theta\)
\(\cos\left(\theta-\dfrac{\pi}{4}\right)=\cos\theta\cos\dfrac{\pi}{4}+\sin\theta\sin\dfrac{\pi}{4}\) \(=\dfrac{\sqrt{2}}{2}\sin\theta+\dfrac{\sqrt{2}}{2}\cos\theta\)
\(\cos\left(\theta-\dfrac{\pi}{3}\right)=\cos\theta\cos\dfrac{\pi}{3}+\sin\theta\sin\dfrac{\pi}{3}\) \(=\dfrac{\sqrt{3}}{2}\sin\theta+\dfrac{1}{2}\cos\theta\)
\(\cos\left(\theta-\dfrac{\pi}{2}\right)=\cos\theta\cos\dfrac{\pi}{2}+\sin\theta\sin\dfrac{\pi}{2}\) \(=\sin\theta\)

\(\sin\theta+\cos\theta=\sqrt{2}\sin\left(\theta+\dfrac{\pi}{4}\right)=\sqrt{2}\cos\left(\theta-\dfrac{\pi}{4}\right)\)
\(\sin\theta+\sqrt{3}\cos\theta=2\sin\left(\theta+\dfrac{\pi}{3}\right)=2\cos\left(\theta-\dfrac{\pi}{6}\right)\)
\(\sqrt{3}\sin\theta-\cos\theta=2\sin\left(\theta-\dfrac{\pi}{6}\right)=2\cos\left(\theta+\dfrac{4\pi}{3}\right)\)

実際に，
\(r\sin(\theta+\alpha)=r(\sin\theta\cos\alpha+\cos\theta\sin\alpha)\) \(=r\left(\dfrac{a}{r}\sin\theta+\dfrac{b}{r}\cos\theta\right)=a\sin\theta+b\cos\theta\)