\(\displaystyle{ -cos(2t)=cos(t)}\)
\(\displaystyle{ ctg( \frac{t}{2})=tg(t)}\)
\(\displaystyle{ ctg(t- \frac{pi}{6} )=tg(t)}\)
równania,cosinusy,tangensy itp.
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Andreas
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- Posty: 1130
- Rejestracja: 1 lis 2008, o 22:33
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równania,cosinusy,tangensy itp.
\(\displaystyle{ -\cos(2t)=\cos(t) \\
- \left(2 \cos^2(t) -1 \right) =\cos(t) \\
2 \cos^2 (t)+ \cos(t)-1=0 \\
x =\cos(t) \\
2x^2+x-1=0 \\
\Delta=1+4 \cdot 2=9 \\
x_1=-1, \ \ x_2= \frac{1}{2} \\
\cos(t)=-1 \vee \cos(t)=\frac{1}{2} \\
t=\pi + 2k\pi \vee t= \frac{1}{3}\pi +2k\pi \vee t= \frac{5}{3} \pi +2k\pi, k \in C}\)
\(\displaystyle{ \ctg \left( \frac{t}{2} \right) =\tg(t) \\
\frac{\sin(t)}{1-\cos(t)} = \frac{\sin(t)}{\cos(t)} \\
1-\cos(t)=\cos(t) \\
2\cos(t)=1 \\
\cos(t)= \frac{1}{2} \\
t= \frac{\pi}{3} +2k\pi \vee t=\frac{-\pi}{3} +2k\pi, k \in C}\)
- \left(2 \cos^2(t) -1 \right) =\cos(t) \\
2 \cos^2 (t)+ \cos(t)-1=0 \\
x =\cos(t) \\
2x^2+x-1=0 \\
\Delta=1+4 \cdot 2=9 \\
x_1=-1, \ \ x_2= \frac{1}{2} \\
\cos(t)=-1 \vee \cos(t)=\frac{1}{2} \\
t=\pi + 2k\pi \vee t= \frac{1}{3}\pi +2k\pi \vee t= \frac{5}{3} \pi +2k\pi, k \in C}\)
\(\displaystyle{ \ctg \left( \frac{t}{2} \right) =\tg(t) \\
\frac{\sin(t)}{1-\cos(t)} = \frac{\sin(t)}{\cos(t)} \\
1-\cos(t)=\cos(t) \\
2\cos(t)=1 \\
\cos(t)= \frac{1}{2} \\
t= \frac{\pi}{3} +2k\pi \vee t=\frac{-\pi}{3} +2k\pi, k \in C}\)