postać trygonometryczna

[Wednesday]

Przedstaw w postaci trygonometrycznej:

1) \(\displaystyle{ 2+ \sqrt{3}+i}\)

2) \(\displaystyle{ \frac{1+i tgx}{1-i tgx}}\)

3) \(\displaystyle{ 1+cosx+isinx}\)

[dabros]

a dwa ostatnie wzory to nib co jest....?

[Wasilewski]

1)
\(\displaystyle{ a = 2 + \sqrt{3} \ \ b = 1 \\
z = |z|(\frac{a}{|z|} + i\frac{b}{|z|}) \\
\frac{a}{|z|} = cos\phi \ \ \frac{b}{|z|} = sin\phi \\
z = \sqrt{6 + 4\sqrt{3}}(cos\phi + isin\phi)}\)

[Rogal]

1)
\(\displaystyle{ |z| = \sqrt{7+4\sqrt{3}+1} = \sqrt{8+4\sqrt{3}} = 2\sqrt{2+\sqrt{3}} \\ \cos \phi = \frac{2+\sqrt{3}}{2\sqrt{2+\sqrt{3}}} = \frac{\sqrt{2+\sqrt{3}}}{2} \\ \sin \phi = \frac{1}{2\sqrt{2+\sqrt{3}}} = \frac{\sqrt{2-\sqrt{3}}}{2} \\ \phi = \frac{\pi}{12} \\ z = 2\sqrt{2+\sqrt{3}}(\cos \frac{\pi}{12} + i\sin \frac{\pi}{12})}\)

[Wasilewski]

O kurczę, chyba już liczyć nie umiem

[Lukasz_C747]

a dwa ostatnie wzory to nib co jest....?

To nie jest postać trygnometryczna. Napisano wyżej jak ona wygląda:
\(\displaystyle{ r(\cos{\phi}+i\sin{\phi})}\)

\(\displaystyle{ b) \frac{1+i\tan{x}}{1-i\tan{x}} = \frac{1+i\frac{\sin{x}}{\cos{x}}}{1-i\frac{\sin{x}}{\cos{x}}} = \frac{\cos{x}+i\sin{x}}{\cos{x}-i\sin{x}} = (\frac{\cos{x}+i\sin{x}}{\cos{x}-i\sin{x}})*(\frac{\cos{x}+i\sin{x}}{\cos{x}+i\sin{x}}) = \frac{(\cos{x}+i\sin{x})^2}{(\cos{x})^2-(i\sin{x})^2} = \frac{(\cos{x}+i\sin{x})^2}{(\cos{x})^2+(\sin{x})^2} = (\cos{x}+i\sin{x})^2 = (\cos{x})^2+2i\cos{x}\sin{x}+(i\sin{x})^2 = (\cos{x})^2-(\sin{x})^2+i\sin{2x} = \cos{2x}+i\sin{2x}}\)

[Qń]

Ad. 3
\(\displaystyle{ 1+ \cos x = 2 \cos^2 \frac{x}{2} \\
\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}}\)
stąd
\(\displaystyle{ 1+ \cos x + i \sin x = 2 \cos^2 \frac{x}{2} + 2isin \frac{x}{2} \cos \frac{x}{2} =
2cos \frac{x}{2} (cos \frac{x}{2} + i \sin \frac{x}{2})}\)

Pozdrawiam.
Qń.