Упр.23.22 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

Задача

Упростите выражение:

1) $$\frac{\sin(3a)}{\sin(a)}-\frac{\cos(3a)}{\cos(a)}$$;

5) $$\frac{4\operatorname{tg}(a)(1-\operatorname{tg}^2(a))}{(1+\operatorname{tg}^2(a))^2}$$;

2) $$\frac{1}{\operatorname{ctg}\left(\frac{a}{2}\right)-\operatorname{tg}\left(\frac{?}{2}\right)}$$;

6) $$\frac{\operatorname{tg}(a)\operatorname{tg}(a)}{\operatorname{tg}(2a)-\operatorname{tg}(a)}$$;

3) $$(\operatorname{tg}(a)+\operatorname{ctg}(a))\sin(2a)$$;

7) $$\frac{\sin^4(a)-\cos^4(a)+\cos^2(a)}{2(1-\cos(a))}$$;

4) $$\frac{1}{1-\operatorname{tg}(a)}-\frac{1}{1+\operatorname{tg}(a)}$$;

8) $$2\cos\left(\frac{?}{4}-\frac{a}{2}\right)\cos\left(\frac{?}{4}+\frac{a}{2}\right)$$.

Подробный ответ

1) $$\frac{\sin 3a}{\sin a}-\frac{\cos 3a}{\cos a}=\frac{\sin 3a\cdot \cos a-\cos 3a\cdot \sin a}{\sin a\cdot \cos a}=\frac{\sin(3a-a)}{\frac12\sin 2a}=\frac{\sin 2a}{\frac12\sin 2a}=2.$$

2) $$\frac{1}{\ctg \frac a2-\tg \frac a2}=\frac{1}{\frac{\cos \frac a2}{\sin \frac a2}-\frac{\sin \frac a2}{\cos \frac a2}}=\frac{\sin \frac a2\cdot \cos \frac a2}{\cos^2 \frac a2-\sin^2 \frac a2}=\frac{\frac12\sin a}{\cos a}=\frac12\tg a.$$

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

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

5) $$\frac{4\tg a\left(1-\tg^2 a\right)}{\left(1+\tg^2 a\right)^2}=2\cdot \frac{2\tg a}{1+\tg^2 a}\cdot \frac{1-\tg^2 a}{1+\tg^2 a}=2\sin 2a\cdot \cos 2a=\sin 4a.$$

6) $$\frac{\tg 2a\cdot \tg a}{\tg 2a-\tg a}=\frac{\frac{2\tg a}{1-\tg^2 a}\cdot \tg a}{\frac{2\tg a}{1-\tg^2 a}-\tg a}=\frac{2\tg^2 a}{2\tg a-\tg a\left(1-\tg^2 a\right)}=\frac{2\tg a}{1+\tg^2 a}=\sin 2a.$$

7) $$\frac{\sin^4 a-\cos^4 a+\cos^2 a}{2(1-\cos a)}=\frac{(1-\cos^2 a)^2-\cos^4 a+\cos^2 a}{2(1-\cos a)}=\frac{1-2\cos^2 a+\cos^4 a-\cos^4 a+\cos^2 a}{2(1-\cos a)}$$

$$=\frac{1-\cos^2 a}{2(1-\cos a)}=\frac{(1-\cos a)(1+\cos a)}{2(1-\cos a)}=\frac{1+\cos a}{2}=\cos^2 \frac a2.$$

8) $$2\cos\left(\frac{\pi}{4}-\frac a2\right)\cos\left(\frac{\pi}{4}+\frac a2\right)=2\cos\left(\frac{\pi}{2}-\left(\frac{\pi}{4}+\frac a2\right)\right)\cos\left(\frac{\pi}{4}+\frac a2\right)$$

$$=2\sin\left(\frac{\pi}{4}+\frac a2\right)\cos\left(\frac{\pi}{4}+\frac a2\right)=\sin\left(\frac{\pi}{2}+a\right)=\cos a.$$