Question

2)Найдите значение выражения 1) sin225*. 2)cos13п/6.3)tg5п/3 4) cos^2(7п/3)-sin^2(7п/3)

3)докажите тождество

4)упростите выражение

Answer 1

$2)\\sin225=sin(180+45)=-sin45=-\frac{\sqrt2}{2}\\cos\frac{13\pi}{6}=cos(2\pi+\frac{\pi}{6})=cos\frac{\pi}{6}=\frac{\sqrt3}{2}\\tg\frac{5\pi}{3}=tg(2\pi-\frac{\pi}{3})=-tg\frac{\pi}{3}=-\sqrt3\\cos^2\frac{7\pi}{3}-sin^2\frac{7\pi}{3}=cos\frac{14\pi}{3}=cos(5\pi-\frac{\pi}{3})=-cos\frac{\pi}{3}=-\frac{1}{2}\\$

$3)\\\frac{sina}{1-cosa}=\frac{1+cosa}{sina}\\\frac{sina}{2sin^2\frac{a}{2}}=\frac{2cos^2\frac{a}{2}}{sina}\\\frac{2sin\frac{a}{2}cos\frac{a}{2}}{2sin^2\frac{a}{2}}=\frac{2cos^2\frac{a}{2}}{2sin\frac{a}{2}cos\frac{a}{2}}\\\frac{cos\frac{a}{2}}{sin\frac{a}{2}}=\frac{cos\frac{a}{2}}{sin\frac{a}{2}}\\ctg\frac{a}{2}=ctg\frac{a}{2}$

$4)\\\frac{1}{tga+ctga}=\frac{1}{\frac{sina}{cosa}+\frac{cosa}{sina}}=\frac{1}{\frac{sin^2a+cos^2a}{sinacosa}}=\frac{1}{\frac{1}{sinacosa}}=sinacosa=\frac{1}{2}sin2a\\cos5\beta cos2\beta + sin5 \beta sin 2\beta =cos(5 \beta -2 \beta )=cos 3\beta\\\frac{2cos^2atga}{tg2a}=\frac{2cos^2a\frac{sina}{cosa}}{tg2a}=\frac{2sinacosa}{\frac{sin2a}{cos2a}}=\frac{sin2a}{{\frac{sin2a}{cos2a}}}=cos2a$