Question

Докажите равенство sin 2 альфа = 2 tg альфа/ 1+ tg в квадрате альфа

Answer 1

sin 2α = 2sinα*cosα

2tgα / 1 + tg²α = 2tgα*cos²α = 2cosα*sinα

2sinα*cosα = 2cosα*sinα

sin 2α = sin 2α

Доказано.

Answer 2

$<var>sin^{2}\alpha=\frac{2tg\alpha}{1+tg^{2}\alpha}, \frac{2tg\alpha}{1+tg^{2}\alpha}-sin^{2}\alpha= </var>$$<var>\frac{2tg\alpha}{\frac{1}{cos^{2}\alpha}}-sin^{2}\alpha=2tg\alpha * cos^{2}\alpha-sin^{2}\alpha= \frac{2sin\alpha}{cos\alpha} * cos^{2}\alpha-sin^{2}\alpha= 2sin\alpha- sin^{2}\alpha=sin^{2}\alpha.</var>$