1)
\(\displaystyle{ sin^{2}x-cos^{2}x=\frac{tx^2x-1}{tg^2x+1}}\)
2)
\(\displaystyle{ \frac{1}{sinx}+ctgx=\frac{sinx}{1-cosx}}\)
2 proste (?) tożsamości
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- Rejestracja: 10 gru 2007, o 20:10
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2 proste (?) tożsamości
1)
\(\displaystyle{ L = sin^{2}x - cos^{2}x = \frac{sin^{2}x - cos^{2}x}{sin^{2}x + cos^{2}x} = \frac{cos^{2}x(\frac{sin^{2}x}{cos^{2}x} - 1)}{cos^{2}x(\frac{sin^{2}x}{cos^{2}x} + 1)} = \frac{tg^{2}x - 1}{tg^{2}x + 1} = P}\)
2)
\(\displaystyle{ P = \frac{sinx}{1 - cosx}\cdot \frac{1 + cosx}{1 + cosx} = \frac{sinx(1 + cosx)}{1 - cos^{2}x} = \frac{sinx(1 + cosx)}{sin^{2}x} = \frac{1 + cosx}{sinx} = \frac{1}{sinx} + ctgx = L}\)
\(\displaystyle{ L = sin^{2}x - cos^{2}x = \frac{sin^{2}x - cos^{2}x}{sin^{2}x + cos^{2}x} = \frac{cos^{2}x(\frac{sin^{2}x}{cos^{2}x} - 1)}{cos^{2}x(\frac{sin^{2}x}{cos^{2}x} + 1)} = \frac{tg^{2}x - 1}{tg^{2}x + 1} = P}\)
2)
\(\displaystyle{ P = \frac{sinx}{1 - cosx}\cdot \frac{1 + cosx}{1 + cosx} = \frac{sinx(1 + cosx)}{1 - cos^{2}x} = \frac{sinx(1 + cosx)}{sin^{2}x} = \frac{1 + cosx}{sinx} = \frac{1}{sinx} + ctgx = L}\)