Question

Буду доказать тождество: (4sinacosa)/(cos^2a-sin^2a)=2tg2a

Answer 1

$\frac{4\sin \alpha \cos \alpha}{\cos^2\alpha-\sin^2\alpha}=2tg2\alpha\ \ |(:2)\\\\\\ \frac{2\sin\alpha\cos\alpha}{\cos^2\alpha-\sin^2\alpha}= \frac{2}{ctg\alpha-tg\alpha} \\\\\\ \frac{2\sin\alpha\cos\alpha}{\cos^2\alpha-\sin^2\alpha}= \frac{2}{ \frac{\cos\alpha}{\sin\alpha}- \frac{\sin\alpha}{\cos\alpha}}\\\\\\ \frac{2\sin\alpha\cos\alpha}{\cos^2\alpha-\sin^2\alpha}= \frac{2}{ \frac{\cos^2\alpha-\sin^2\alpha}{\sin\alpha\cos\alpha}}\\\\\\ \frac{2\sin\alpha\cos\alpha}{\cos^2\alpha-\sin^2\alpha}=\frac{2\sin\alpha\cos\alpha}{\cos^2\alpha-\sin^2\alpha}$

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

Можно доказать тождество проще:

$\frac{4\sin\alpha\cos\alpha}{\cos^2\alpha-\sin^2\alpha}=2tg2\alpha\\\\ \frac{2\sin2\alpha}{\cos2\alpha}=2tg2\alpha\\\\ 2tg2\alpha= 2tg2\alpha$