Rozwiązanie:
Mamy:
[tex]sinx+cosx=\frac{\sqrt{5} }{5}[/tex]
[tex]a)[/tex]
[tex]sinxcosx= \ ?[/tex]
Zauważmy, że:
[tex](sinx+cosx)^{2}=sin^{2}x+2sinxcosx+cos^{2}x=1+2sinxcosx=\frac{1 }{5} \\2sinxcosx=\frac{1 }{5}-1=-\frac{4}{5} \\sinxcosx=-\frac{2}{5}[/tex]
[tex]b)[/tex]
[tex]sin^{4}x+cos^{4}x=\ ?[/tex]
Zauważmy, że:
[tex]sin^{4}x+cos^{4}x=(sin^{2}x)^{2}+(cos^{2}x)^{2}=(sin^{2}x+cos^{2}x)^{2}-2(sinxcosx)^{2}=1-2(sinxcosx)^{2}=1-2 \cdot (-\frac{2}{5})^{2} =\frac{17}{25}[/tex]
[tex]c)[/tex]
[tex]sin^{3}x+cos^{3}x= \ ?[/tex]
Zauważmy, że:
[tex]sin^{3}x+cos^{3}x=(sinx+cosx)(sin^{2}x-sinxcosx+cos^{3}x)=\frac{\sqrt{5} }{5} (1+\frac{2}{5})= \frac{\sqrt{5} }{5} \cdot \frac{7}{5} =\frac{7\sqrt{5} }{25}[/tex]
[tex]d)[/tex]
[tex]|sinx-cosx|= \ ?[/tex]
Zauważmy, że:
[tex]|sinx-cosx|=\sqrt{(sinx-cosx)^{2}} =\sqrt{sin^{2}x-2sinxcosx+cos^{2}x} =\sqrt{1-2sinxcosx} =\sqrt{1-2 \cdot (-\frac{2}{5} )} =\sqrt{\frac{9}{5} } =\frac{3\sqrt{5} }{5}[/tex]
