Question

Помогите пожалуйста людиииии ​

Answer 1

Ответ:

$1)\ \ 1500^\circ =\dfrac{1500^\circ \cdot \pi }{180^\circ }=\dfrac{50\pi }{6}=\dfrac{25\pi }{3}\\\\360^\circ =2\pi \\\\2400^\circ =\dfrac{2400^\circ \cdot \pi }{180^\circ }=\dfrac{40\pi }{3}\\\\3290^\circ =\dfrac{3200^\circ \cdot \pi }{180^\circ }=\dfrac{16\pi }{9}$

$2)\ \ \dfrac{sin(\pi -a)-cos(\frac{\pi}{2}+a)}{cos(\pi +a)}=\dfrac{sina+sina}{-cosa}=-\dfrac{2sina}{cosa}\\\\\\3)\ \ cos70^\circ \cdot cos40^\circ +sin70^\circ \cdot sin40^\circ =cos(70^\circ -40^\circ )=cos30^\circ =\dfrac{\sqrt3}{2}\\\\\\sin75^\circ =sin(45^\circ +30^\circ )=sin45^\circ \cdot cos30^\circ +cos45^\circ \cdot sin30^\circ =\\\\=\dfrac{\sqrt2}{2}\cdot \dfrac{\sqrt3}{2}+\dfrac{\sqrt2}{2}\cdot \dfrac{1}{2}=\dfrac{\sqrt6+\sqrt2}{4}$

$4)\ \ \dfrac{sin2a}{2ctga}=\dfrac{2\, sina\cdot cosa}{2\cdot \frac{cosa}{sina}}=\dfrac{sin^2a\cdot cosa}{cosa}=sin^2a\\\\\\\dfrac{1-cos2a}{1+cos2a}=\dfrac{2sin^2a}{2cos^2a}=tg^2a$