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kokokosek@wp.pl · Post autor: **kokokosek@wp.pl** » 8 kwie 2009, o 21:29

\(\displaystyle{ \frac{cos ^{3}x-sin^{3}x }{cosx-sinx}= \frac{2+sin2x}{2}}\)

\(\displaystyle{ \frac{1}{sin^{2}x}+ \frac{1}{cos^{2}x}= \frac{1+ctg^{2}x}{cos^{2}x}}\)

\(\displaystyle{ sin^{2}x-cos^{2}x= \frac{tg^{2}x-1}{tg^{2}x+1}}\)

lukki_173 · Post autor: **lukki_173** » 8 kwie 2009, o 21:42

\(\displaystyle{ \frac{1}{sin^{2}x}+ \frac{1}{cos^{2}x}= \frac{1+ctg^{2}x}{cos^{2}x} \\
L=\frac{1}{sin^{2}x}+ \frac{1}{cos^{2}x}= \frac{sin^{2}x+cos^{2}x}{sin^{2}xcos^{2}x}= \frac{1}{sin^{2}xcos^{2}x}\\ \\
P=\frac{1+ctg^{2}x}{cos^{2}x}= \frac{ \frac{sin^{2}x}{sin^{2}x}+ \frac{cos^{2}x}{sin^{2}x} }{cos^{2}x}= \frac{ \frac{1}{sin^{2}x} }{cos^{2}x}= \frac{1}{sin^{2}xcos^{2}x}}\)
Zatem:
\(\displaystyle{ L=P}\)
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wb · Post autor: wb » 8 kwie 2009, o 21:42

\(\displaystyle{ L=\frac{cos ^{3}x-sin^{3}x }{cosx-sinx}= \frac{(cosx-sinx)(cos^2x+sinxcosx+sin^2x)}{cosx-sinx}= 1+sinxcosx= \\ \frac{2+2sinxcosx}{2}= \frac{2+sin2x}{2}=P}\)

lukki_173 · Post autor: **lukki_173** » 8 kwie 2009, o 21:51

\(\displaystyle{ sin^{2}x-cos^{2}x= \frac{tg^{2}x-1}{tg^{2}x+1}}\)
\(\displaystyle{ P=\frac{tg^{2}x-1}{tg^{2}x+1}= \frac{ \frac{sin^{2}x-cos^{2}x}{cos^{2}x} }{ \frac{sin^{2}x+cos^{2}x}{cos^{2}x} } = \frac{sin^{2}x-cos^{2}x}{sin^{2}x+cos^{2}x} =sin^{2}x-cos^{2}x=L}\)
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