Упр.20.2 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)
- Упростите выражение:
1) $$\sin^2(2a)+\cos^2(2a)+\operatorname{ctg}^2(5a)$$;
5) $$\bigl(\operatorname{tg}(a)\cos(a)\bigr)^2+\bigl(\operatorname{ctg}(a)\sin(a)\bigr)^2$$;
2) $$\sin\left(\frac{a}{3}\right)\operatorname{ctg}\left(\frac{a}{3}\right)$$;
6) $$\frac{\sin^2(a)}{1+\operatorname{ctg}^2(a)\left(\cos^2(a)-1\right)}$$;
3) $$1-\frac{1}{\sin^2(?)}$$;
7) $$\left(\frac{1}{\cos(a)}+\operatorname{tg}(a)\right)\left(\frac{1}{\cos(a)}-\operatorname{tg}(a)\right)$$;
4) $$\frac{\sin^2(a)-1}{\cos^2(a)-1}+\operatorname{tg}(a)\operatorname{ctg}(a)$$;
8) $$\left(\operatorname{tg}(?)+\operatorname{ctg}(?)\right)^2-\left(\operatorname{tg}(?)-\operatorname{ctg}(?)\right)^2$$.
$$\sin^2 2a+\cos^2 2a+\ctg^2 5a=1+\ctg^2 5a.$$
По формуле $$1+\ctg^2 x=\frac{1}{\sin^2 x}$$ получаем
$$1+\ctg^2 5a=\frac{1}{\sin^2 5a}.$$
$$\sin \frac{a}{3}\cdot \ctg \frac{a}{3}=\sin \frac{a}{3}\cdot \frac{\cos \frac{a}{3}}{\sin \frac{a}{3}}=\cos \frac{a}{3}.$$
$$1-\frac{1}{\sin^2 \gamma}=1-\left(1+\ctg^2 \gamma\right)=-\ctg^2 \gamma.$$
$$\frac{\sin^2 a-1}{\cos^2 a-1}+\tg a\cdot \ctg a$$
$$=\frac{\sin^2 a-(\sin^2 a+\cos^2 a)}{\cos^2 a-(\sin^2 a+\cos^2 a)}+1$$
$$=\frac{-\cos^2 a}{-\sin^2 a}+1=\ctg^2 a+1=\frac{1}{\sin^2 a}.$$
$$\left(\tg a\cdot \cos a\right)^2+\left(\ctg a\cdot \sin a\right)^2$$
$$=\left(\frac{\sin a}{\cos a}\cdot \cos a\right)^2+\left(\frac{\cos a}{\sin a}\cdot \sin a\right)^2$$
$$=\sin^2 a+\cos^2 a=1.$$
$$\frac{\sin^2 a}{1+\ctg^2 a\cdot (\cos^2 a-1)}$$
$$=\frac{\sin^2 a}{1+\ctg^2 a\cdot \bigl(\cos^2 a-(\sin^2 a+\cos^2 a)\bigr)}$$
$$=\frac{\sin^2 a}{1+\ctg^2 a\cdot (-\sin^2 a)}$$
$$=\frac{\sin^2 a}{1-\frac{\cos^2 a}{\sin^2 a}\cdot \sin^2 a}=\frac{\sin^2 a}{1-\cos^2 a}=\frac{\sin^2 a}{\sin^2 a}=1.$$
$$\left(\frac{1}{\cos a}+\tg a\right)\left(\frac{1}{\cos a}-\tg a\right)$$
$$=\frac{1}{\cos^2 a}-\tg^2 a=(1+\tg^2 a)-\tg^2 a=1.$$
$$\left(\tg \beta+\ctg \beta\right)^2-\left(\tg \beta-\ctg \beta\right)^2$$
$$=\left(\tg^2 \beta+\ctg^2 \beta+2\tg \beta\cdot \ctg \beta\right)-\left(\tg^2 \beta+\ctg^2 \beta-2\tg \beta\cdot \ctg \beta\right)$$
$$=4\tg \beta\cdot \ctg \beta=4.$$
