Oblicz

[Eqauzm]

1.Zapisz w postaci iloczynu \(\displaystyle{ \sin + \tan }\)

2. Rozwiąż równanie \(\displaystyle{ \sin 3x = 0,5}\)

3.Oblicz \(\displaystyle{ \sin 2\alpha}\) jeżeli \(\displaystyle{ \sin = \frac{3}{5}}\)

[wb]

1.
\(\displaystyle{ =sin\alpha+ \frac{sin\alpha}{cos\alpha}=sin\alpha(1+ \frac{1}{cos\alpha})}\)


2.
\(\displaystyle{ 3x= \frac{\pi}{6}+2k\pi 3x= \frac{5\pi}{6}+2k\pi \\ x= \frac{\pi}{18}+ \frac{2k\pi}{3} x= \frac{5\pi}{18}+ \frac{2k\pi}{3} \ \ \ \ ; \ \ \ k\in C}\)


3.
\(\displaystyle{ sin2\alpha = 2sin\alpha cos\alpha \\ \\ sin\alpha= \frac{3}{5} cos\alpha=\pm \sqrt{1- (\frac{3}{5})^2}=\pm \frac{4}{5}}\)

i oblicz \(\displaystyle{ sin2\alpha}\) w dwu przypadkach.

[szablewskil]

Zad 3
Z jedynki trygonometrycznej \(\displaystyle{ cos\alpha= \sqrt{1-sin\alpha^{2}}= \frac{4}{5}}\)
Teraz liczysz tangens alfa \(\displaystyle{ tan\alpha= \frac{sin\alpha}{cos\alpha}= \frac{3}{4}}\)
A teraz \(\displaystyle{ sin2\alpha}\) na podstawie tangensa alfa:
\(\displaystyle{ sin2\alpha= \frac{2tan\alpha}{1+tan\alpha^{2}}= \frac{24}{25}}\)

Mozna latwiej
\(\displaystyle{ sin2\alpha=2*sin\alpha*cos\alpha=2* \frac{3}{5}* \frac{4}{5}= \frac{24}{25}}\)