Упр.8 ГДЗ Колмогоров 10-11 класс (Алгебра)

10–11

Упростим выражения, используя основное тригонометрическое тождество $$\sin^2 x+\cos^2 x=1$$.

1. $$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha$$

   $$\cos^2\alpha-\cos^4\alpha+\sin^4\alpha =\cos^2\alpha+(\sin^2\alpha)^2-(\cos^2\alpha)^2$$
   $$=\cos^2\alpha+(\sin^2\alpha+\cos^2\alpha)(\sin^2\alpha-\cos^2\alpha)$$
   $$=\cos^2\alpha+\sin^2\alpha-\cos^2\alpha=\sin^2\alpha.$$

2. $$\frac{1-2\cos^2\beta}{\cos\beta+\sin\beta}$$

   $$\frac{1-2\cos^2\beta}{\cos\beta+\sin\beta} =\frac{\sin^2\beta+\cos^2\beta-2\cos^2\beta}{\cos\beta+\sin\beta}$$
   $$=\frac{\sin^2\beta-\cos^2\beta}{\cos\beta+\sin\beta} =\frac{(\sin\beta-\cos\beta)(\sin\beta+\cos\beta)}{\cos\beta+\sin\beta}$$
   $$=\sin\beta-\cos\beta.$$

   При этом $$\cos\beta+\sin\beta\neq 0$$, то есть $$\beta\neq -\frac{\pi}{4}+\pi n,\ n\in\mathbb Z.$$

3. $$\left(\sin^2\alpha+\tg^2\alpha\sin^2\alpha\right)\ctg\alpha$$

   $$\left(\sin^2\alpha+\tg^2\alpha\sin^2\alpha\right)\ctg\alpha =\sin^2\alpha\left(1+\tg^2\alpha\right)\ctg\alpha$$
   $$=\sin^2\alpha\cdot\frac{1}{\cos^2\alpha}\cdot\frac{\cos\alpha}{\sin\alpha} =\tg\alpha.$$

4. $$\frac{\sin^2 t-1}{\cos^4 t}+\tg^2 t$$

   $$\frac{\sin^2 t-1}{\cos^4 t}+\tg^2 t =\frac{\sin^2 t-(\sin^2 t+\cos^2 t)}{\cos^4 t}+\frac{\sin^2 t}{\cos^2 t}$$
   $$=\frac{-\cos^2 t}{\cos^4 t}+\frac{\sin^2 t}{\cos^2 t} =-\frac{1}{\cos^2 t}+\frac{\sin^2 t}{\cos^2 t}$$
   $$=\frac{-(\sin^2 t+\cos^2 t)+\sin^2 t}{\cos^2 t} =-\frac{\cos^2 t}{\cos^2 t}=-1.$$

Ответ: а) $$\sin^2\alpha$$; б) $$\sin\beta-\cos\beta$$; в) $$\tg\alpha$$; г) $$-1$$.