Упр.52 Повторение ГДЗ Колмогоров 10-11 класс (Алгебра)
Упростите выражения:
- а) $$\operatorname{tg}^2(?) — \sin^2(?) — \operatorname{tg}^2(?)\sin^2(?)$$; б) $$\sqrt{\sin^2(?)(1+\operatorname{ctg}(?))+\cos^2(?)(1+\operatorname{tg}(?))}$$; в) $$(3\sin(?)+2\cos(?))^2+(2\sin(?)-3\cos(?))^2$$; г) $$\frac{\cos(?)\operatorname{tg}(?)}{\sin^2(?)}-\operatorname{ctg}(?)\cos(?)$$.
а) $$\tg^2\alpha-\sin^2\alpha-\tg^2\alpha\cdot\sin^2\alpha$$
Так как $$\tg^2\alpha=\dfrac{1}{\cos^2\alpha}-1,$$ получаем:
$$\tg^2\alpha-\sin^2\alpha-\tg^2\alpha\sin^2\alpha = \tg^2\alpha-\sin^2\alpha-\left(\dfrac{1}{\cos^2\alpha}-1\right)\sin^2\alpha$$
$$= \tg^2\alpha-\sin^2\alpha-\tg^2\alpha+\sin^2\alpha=0.$$
б) $$\sqrt{\sin^2\beta(1+\ctg\beta)+\cos^2\beta(1+\tg\beta)}$$
Раскроем скобки:
$$\sqrt{\sin^2\beta+\sin\beta\cos\beta+\cos^2\beta+\sin\beta\cos\beta} = \sqrt{(\sin\beta+\cos\beta)^2}$$
$$=|\sin\beta+\cos\beta|.$$
в) $$\left(3\sin\alpha+2\cos\alpha\right)^2+\left(2\sin\alpha-3\cos\alpha\right)^2$$
Раскроем скобки:
$$9\sin^2\alpha+12\sin\alpha\cos\alpha+4\cos^2\alpha + 4\sin^2\alpha-12\sin\alpha\cos\alpha+9\cos^2\alpha$$
$$=13\sin^2\alpha+13\cos^2\alpha =13(\sin^2\alpha+\cos^2\alpha)=13.$$
г) $$\dfrac{\cos\beta\cdot\tg\beta}{\sin^2\beta}-\ctg\beta\cdot\cos\beta$$
Преобразуем:
$$\dfrac{\cos\beta\cdot\tg\beta}{\sin^2\beta}-\ctg\beta\cdot\cos\beta = \dfrac{\cos\beta\cdot\dfrac{\sin\beta}{\cos\beta}}{\sin^2\beta} — \dfrac{\cos\beta\cdot\cos\beta}{\sin\beta}$$
$$= \dfrac{1}{\sin\beta}-\dfrac{\cos^2\beta}{\sin\beta} = \dfrac{1-\cos^2\beta}{\sin\beta} = \dfrac{\sin^2\beta}{\sin\beta} = \sin\beta.$$
Ответ
а) $$0$$; б) $$|\sin\beta+\cos\beta|$$; в) $$13$$; г) $$\sin\beta$$.
