Równania trygonometryczne

[mbanan17]

Majac dane: tgx=3 oblicz
\(\displaystyle{ \frac{3cos ^{2}-sin ^{2}x }{sinx \cdot cosx+cos ^{2}x }}\)

[sea_of_tears]

\(\displaystyle{ tgx=3\newline
tgx=\frac{sinx}{cosx}\newline
\frac{sinx}{cosx}=3\newline
sinx=3cosx\newline
\newline
sin^2x+cos^2x=1\newline
(3cosx)^2+cos^2x=1\newline
9cos^2x+cos^2x=1\newline
10cos^2x=1\newline
cos^2x=\frac{1}{10}\newline
cosx=\sqrt{\frac{1}{10}}\newline
cosx=\frac{\sqrt{10}}{10}\newline
sinx=3\cdot \frac{\sqrt{10}}{10}=\frac{3\sqrt{10}}{10}\newline
\newline
\frac{3cos^2x-sin^2x}{sinx\cdot cosx+cos^2x}=
\frac{3\cdot (\frac{\sqrt{10}}{10})^2-(\frac{3\sqrt{10}}{10})^2}{(\frac{3\sqrt{10}}{10})\cdot (\frac{\sqrt{10}}{10})+(\frac{\sqrt{10}}{10})^2}=
\frac{3\cdot \frac{1}{10}-\frac{9}{10}}{\frac{3}{10}+\frac{1}{10}}=
\frac{-\frac{6}{10}}{\frac{4}{10}}=-\frac{6}{4}=-1,5}\)

[mbanan17]

dzieki;) masz pomysl na to? bo cos nie radze sobie z ta trygonometria:(

1)majac dane
\(\displaystyle{ \frac{6sinx+5cosx}{4sinx+cosx}=2}\)

oblicz cos2x

2)Wiedzac ze
\(\displaystyle{ cosx = \frac{1}{2}}\)
oblicz
\(\displaystyle{ sin ^{2}x-cos ^{2}x}\); \(\displaystyle{ \frac{sin ^{2}x }{1+cosx}}\)

[sea_of_tears]

\(\displaystyle{ cosx=\frac{1}{2}\newline
sin^2x-cos^2x=(1-cos^2x)-cos^2x=
1-2cos^2x=1-2\cdot (\frac{1}{2})^2=1-2\cdot\frac{1}{4}=1-\frac{1}{2}=\frac{1}{2}\newline
\frac{sin^2x}{1+cosx}=\frac{1-cos^2x}{1+cosx}=\frac{(1-cosx)(1+cosx)}{1+cosx}=
1-cosx=1-\frac{1}{2}=\frac{1}{2}}\)-- 9 lutego 2009, 22:48 --\(\displaystyle{ \frac{6sinx+5cosx}{4sinx+cosx}=2\newline
6sinx+5cosx=2(4sinx+cosx)\newline
6sinx+5cosx=8sinx+2cosx\newline
2sinx=3cosx\newline
sinx=\frac{3}{2}cosx\newline
\newline
sin^2x+cos^2x=1\newline
(\frac{3}{2}cosx)^2+cos^2x=1\newline
\frac{9}{4}cos^2x+cos^2x=1\newline
\frac{13}{4}cos^2x=1\newline
cos^2x=\frac{4}{13}\newline
\newline
cos2x=2cos^2x-1=2\cdot \frac{4}{13}-1=\frac{8}{13}-1=-\frac{5}{13}}\)