Question

ТРИГОНОМЕТРИЯ, 100 БАЛЛОВ
Упростите
1) sin^2 a -1;
2) sin^2 2a + cos^2 2a + ctg^2 5a;
3)2sin * (a/3) * ctg * (a/3) - cos * (a/3);
4)(cos^2 a-1)/(sin^2 a-1) + tg a * ctg a;
5)(tg a * cos a)/(1+ctg^2 a)​

Answer 1

$\displaystyle\bf\\1)\\Sin^{2} \alpha -1=Sin^{2} \alpha -(Sin^{2} \alpha +Cos^{2} \alpha )=Sin^{2} \alpha -Sin^{2} \alpha -Cos^{2} \alpha =-Cos^{2} \alpha \\\\2)\\\underbrace{Sin^{2} 2\alpha +Cos^{2} 2\alpha}_{1}+Ctg^{2} 5\alpha =1+Ctg^{2}5\alpha =\frac{1}{Sin^{2} 5\alpha } \\\\3)$

$\displaystyle\bf\\2Sin\frac{\alpha }{3}\cdot Ctg\frac{\alpha }{3}-Cos\frac{\alpha }{3} =2Sin\frac{\alpha }{3} \cdot \frac{Cos\frac{\alpha }{3} }{Sin\frac{\alpha }{3} }-Cos\frac{\alpha }{3} =2Cos\frac{\alpha }{3} -Cos\frac{\alpha }{3} =Cos\frac{\alpha }{3} \\\\4)\\\frac{Cos^{2} \alpha -1}{Sin^{2} \alpha -1}+tg\alpha \cdot Ctg\alpha =\frac{-Sin^{2} \alpha }{-Cos^{2} \alpha } +1=tg^{2} \alpha +1=\frac{1}{Cos^{2} \alpha }$

$\displaystyle\bf\\5)\\\frac{tg\alpha \cdot Cos\alpha }{1+Ctg^{2} \alpha } =\frac{\frac{Sin\alpha }{Cos\alpha } \cdot Cos\alpha }{\frac{1}{Sin^{2} \alpha } } =Sin\alpha \cdot Sin^{2}\alpha =Sin^{3} \alpha$