Gemma scrive: trigonometria 1

$sen(\alpha) \cdot tg \frac{\alpha}{2}=sen^2 \alpha - 2cos \alpha sen^2 \frac{\alpha}{2}$

Sapendo che:

$sen \alpha= \frac {2tg \frac{\alpha}{2}}{1+tg^2 \frac{\alpha}{2}}$

$sen^2 \frac{\alpha}{2}=\frac{1-cos \alpha}{2}$

avremo:

$\frac {2tg \frac{\alpha}{2}}{1+tg^2 \frac{\alpha}{2}} \cdot tg \frac{\alpha}{2}=sen^2 \alpha - 2cos \alpha \cdot \frac{1-cos \alpha}{2}$

$\frac {2tg^2 \frac{\alpha}{2}}{1+tg^2 \frac{\alpha}{2}}=sen^2 \alpha - cos \alpha +cos^2 \alpha$

$\frac {2tg^2 \frac{\alpha}{2}}{1+tg^2 \frac{\alpha}{2}}=1 - cos \alpha$

Sappiamo che:

$cos \alpha= \frac {1-tg \frac{\alpha}{2}}{1+tg^2 \frac{\alpha}{2}}$

avremo:

$\frac {2tg^2 \frac{\alpha}{2}}{1+tg^2 \frac{\alpha}{2}}=1 - \frac {1-tg \frac{\alpha}{2}}{1+tg^2 \frac{\alpha}{2}}$

$\frac {2tg^2 \frac{\alpha}{2}}{1+tg^2 \frac{\alpha}{2}}= \frac {1+tg \frac{\alpha}{2}-1+tg \frac{\alpha}{2}}{1+tg^2 \frac{\alpha}{2}}$

$\frac {2tg^2 \frac{\alpha}{2}}{1+tg^2 \frac{\alpha}{2}}=\frac {2tg^2 \frac{\alpha}{2}}{1+tg^2 \frac{\alpha}{2}}$

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