Trigonometrie
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Lernkarten
\(\sin(0^{\circ})=\sin(180^{\circ})=\sin(360^{\circ})\)
\(\sin(0)=\sin(\pi)=\sin(2\pi)=0\)
\(\sin(15^{\circ})=\sin(165^{\circ})\\=-\sin(195^{\circ})=-\sin(345^{\circ})\)
\(\sin({\pi\over12})=\sin({11\pi\over12})={\sqrt{6}-\sqrt{2}\over 4}\\ =-\sin({13\pi\over12})=-\sin({23\pi\over12})\)
\(\sin(30^{\circ})=\sin(150^{\circ})\\ =-\sin(210^{\circ})=-\sin(330^{\circ})\)
\(\sin({\pi\over6})=\sin({5\pi\over6})={1\over2}\\ =-\sin({7\pi\over6})=-\sin({11\pi\over6})\)
\(\sin(45^{\circ})=\sin(135^{\circ})\\ =-\sin(225^{\circ})=-\sin(315^{\circ})\)
\(\sin({\pi\over4})=\sin({3\pi\over4})={\sqrt{2}\over2}\\= -\sin({5\pi\over4})=-\sin({7\pi\over4})\)
\(\sin(60^{\circ})=\sin(120^{\circ})\\= -\sin(240^{\circ})=-\sin(300^{\circ})\)
\(\sin({\pi\over3})=\sin({2\pi\over3})={\sqrt{3}\over2}\\= -\sin({4\pi\over3})=-\sin({5\pi\over3})\)
\(\sin(75^{\circ})=\sin(105^{\circ})\\ =-\sin(255^{\circ})=-\sin(285^{\circ})\)
\(\sin({5\pi\over12})=\sin({7\pi\over12})={\sqrt{6}+\sqrt{2}\over4}\\= -\sin({17\pi\over12})=-\sin({19\pi\over12})\)
\(\sin(90^{\circ})=-\sin(270^{\circ})\)
\(\sin({\pi\over2})=\sin({3\pi\over2})=1\)
\(\cos(0^{\circ})=-\cos(180^{\circ})=\cos(360^{\circ})\)
\(\cos(0)=\cos(2\pi)=1 \\ =-\cos(\pi)\)
\(\cos(15^{\circ})=\cos(345^{\circ})\\ =-\cos(165^{\circ})=-\cos(195^{\circ})\)
\(\cos({1\over12}\pi)=\cos({23\over12}\pi)={\sqrt{6}+\sqrt{2}\over4}\\ =-\cos({11\over12}\pi)=-\cos({13\over12}\pi)\)
\(\cos(30^{\circ})=\cos(330^{\circ})\\ =-\cos(150^{\circ})=-\cos(210^{\circ})\)
\(\cos({1\over6}\pi)=\cos({11\over6}\pi)={\sqrt{3}\over2}\\ =-\cos({5\over6}\pi)=-\cos({7\over6}\pi)\)
\(\cos(45^{\circ})=\cos(315^{\circ})\\ =-\cos(135^{\circ})=-\cos(225^{\circ})\)
\(\cos({1\over4}\pi)=\cos({7\over4}\pi)={\sqrt{2}\over2}\\ =-\cos({3\over4}\pi)=-\cos({5\over4}\pi)\)
\(\cos(60^{\circ})=\cos(300^{\circ})\\ =-\cos(120^{\circ})=-\cos(240^{\circ})\)
\(\cos({1\over3}\pi)=\cos({5\over3}\pi)={1\over2}\\ =-\cos({2\over3}\pi)=-\cos({4\over3}\pi)\)
\(\cos(75^{\circ})=\cos(285^{\circ})\\ =-\cos(105^{\circ})=-\cos(255^{\circ})\)
\(\cos({5\over12}\pi)=\cos({19\over12}\pi)={\sqrt{6}-\sqrt{2}\over4}\\ =-\cos({7\over12}\pi)=-\cos({17\over12}\pi)\)
\(\cos(90^{\circ})=\cos(270^{\circ})\)
\(\cos({1\over2}\pi)=\cos({3\over2}\pi)=0\)
