\(sin(arcsin\frac{3}{5}+arcsin\frac{8}{17})\\
sin(arctg1+arctg2)\)
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octahedron
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- Rejestracja: 19 mar 2011, 00:22
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Pierwsze ok
\(\mbox{ctg}^2x=\frac{\cos^2x}{\sin^2x}=\frac{1-\sin^2x}{\sin^2x}=\frac{1}{\sin^2x}-1
\frac{1}{\sin^2x}=1+\mbox{ctg}^2x
\sin x=\frac{1}{\sqrt{1+\mbox{ctg}^2x}}=\frac{1}{\sqrt{1+\frac{1}{\mbox{tg}^2x}}}
\sin(\mbox{arctg}x)=\frac{1}{\sqrt{1+\frac{1}{x^2}}}
\mbox{tg}^2x=\frac{\sin^2x}{\cos^2x}=\frac{1-\cos^2x}{\cos^2x}=\frac{1}{\cos^2x}-1
\cos x=\frac{1}{\sqrt{1+\mbox{tg}^2x}}
\cos(\mbox{arctg}x)=\frac{1}{\sqrt{1+x^2}}
\sin(\mbox{arctg}1+\mbox{arctg}2)=\sin(\frac{\pi}{4}+\mbox{arctg}2)=\sin\frac{\pi}{4}\cos(\mbox{arctg}2)+\sin(\mbox{arctg}2)\cos\frac{\pi}{4}=
=\frac{\sqrt{2}}{2}\(\cos(\mbox{arctg}2)+\sin(\mbox{arctg}2)\)=\frac{\sqrt{2}}{2}\(\cos(\mbox{arctg}2)+\sin(\mbox{arctg}2)\)=
=\frac{\sqrt{2}}{2}\(\frac{1}{\sqrt{1+2^2}}+\frac{1}{\sqrt{1+\frac{1}{2^2}}}\)=\frac{\sqrt{2}}{2}\(\frac{1}{\sqrt{5}}+\frac{2}{\sqrt{5}}\)=\frac{3\sqrt{10}}{10}\)
\(\mbox{ctg}^2x=\frac{\cos^2x}{\sin^2x}=\frac{1-\sin^2x}{\sin^2x}=\frac{1}{\sin^2x}-1
\frac{1}{\sin^2x}=1+\mbox{ctg}^2x
\sin x=\frac{1}{\sqrt{1+\mbox{ctg}^2x}}=\frac{1}{\sqrt{1+\frac{1}{\mbox{tg}^2x}}}
\sin(\mbox{arctg}x)=\frac{1}{\sqrt{1+\frac{1}{x^2}}}
\mbox{tg}^2x=\frac{\sin^2x}{\cos^2x}=\frac{1-\cos^2x}{\cos^2x}=\frac{1}{\cos^2x}-1
\cos x=\frac{1}{\sqrt{1+\mbox{tg}^2x}}
\cos(\mbox{arctg}x)=\frac{1}{\sqrt{1+x^2}}
\sin(\mbox{arctg}1+\mbox{arctg}2)=\sin(\frac{\pi}{4}+\mbox{arctg}2)=\sin\frac{\pi}{4}\cos(\mbox{arctg}2)+\sin(\mbox{arctg}2)\cos\frac{\pi}{4}=
=\frac{\sqrt{2}}{2}\(\cos(\mbox{arctg}2)+\sin(\mbox{arctg}2)\)=\frac{\sqrt{2}}{2}\(\cos(\mbox{arctg}2)+\sin(\mbox{arctg}2)\)=
=\frac{\sqrt{2}}{2}\(\frac{1}{\sqrt{1+2^2}}+\frac{1}{\sqrt{1+\frac{1}{2^2}}}\)=\frac{\sqrt{2}}{2}\(\frac{1}{\sqrt{5}}+\frac{2}{\sqrt{5}}\)=\frac{3\sqrt{10}}{10}\)