Question

Решите 2 примера 3-4sin^2α ; 3-√3tgα .

Answer 1

$3-4sin^2 \alpha =3-3sin^2 \alpha -sin^2 \alpha =3(1-sin^2 \alpha )-sin^2 \alpha =\\\\=3cos^2 \alpha -sin^2 \alpha =2cos^2 \alpha +(cos^2 \alpha -sin^2 \alpha )=2cos^2 \alpha +cos2 \alpha \; ;\\\\\\3-\sqrt3tg \alpha =\sqrt3\cdot (\sqrt3-tg \alpha )=\sqrt3\cdot (tg\frac{\pi}{3}-tg \alpha )=\\\\=\sqrt3\cdot (\frac{sin\frac{\pi}{3}}{cos\frac{\pi}{3}}-\frac{sin \alpha }{cos \alpha })=\sqrt3\cdot \frac{sin\frac{\pi}{3}cos \alpha -sin \alpha cos\frac{\pi}{3}}{cos\frac{\pi}{3}\cdot cos \alpha }=$

$=\sqrt3\cdot \frac{sin( \frac{\pi}{3}-\alpha )}{\frac{1}{2}\cdot cos \alpha }=2\sqrt3\cdot \frac{sin( \frac{\pi}{3}-\alpha )}{cos \alpha }$