Упр.26.13 ГДЗ Мордковича 10 класс профильный уровень (Алгебра)
- Докажите тождество:
а) $$\frac{\operatorname{tg}(\pi-t)}{\cos(\pi+t)}\cdot\frac{\sin\left(\frac{3\pi}{2}+t\right)}{\operatorname{tg}\left(\frac{3\pi}{2}-t\right)}=\operatorname{tg}^2 t;$$
б) $$\frac{\sin(\pi-t)}{\operatorname{tg}(\pi+t)}\cdot\frac{\operatorname{ctg}\left(\frac{\pi}{2}-t\right)}{\operatorname{tg}\left(\frac{\pi}{2}+t\right)}\cdot\frac{\cos(2\pi-t)}{\sin(-t)}=\sin t;$$
в) $$\frac{\cos^2(\pi-t)+\sin^2\left(\frac{\pi}{2}-t\right)+\cos(\pi+t)\cos(2\pi-t)}{\operatorname{tg}^2\left(t-\frac{\pi}{2}\right)\operatorname{ctg}^2\left(\frac{3\pi}{2}+t\right)};$$
г) $$\frac{\sin^2\left(t-\frac{3\pi}{2}\right)\cos(2\pi-t)}{\operatorname{tg}^2\left(t-\frac{\pi}{2}\right)\cos^2\left(t-\frac{3\pi}{2}\right)}=\cos t.$$
а)
$$\frac{\tg(\pi-t)}{\cos(\pi+t)}\cdot \frac{\sin\left(\frac{3\pi}{2}+t\right)}{\tg\left(\frac{3\pi}{2}+t\right)} = \frac{-\tg t}{-\cos t}\cdot \frac{-\cos t}{-\ctg t} = \frac{\tg t}{\ctg t} = \tg t\cdot \frac{1}{\tg t} = \tg^2 t.$$
Тождество доказано.
б)
$$\frac{\sin(\pi-t)}{\tg(\pi+t)}\cdot \frac{\ctg\left(\frac{\pi}{2}-t\right)}{\tg\left(\frac{\pi}{2}+t\right)}\cdot \frac{\cos(2\pi-t)}{\sin(-t)} = \frac{\sin t}{\tg t}\cdot \frac{\tg t}{-\ctg t}\cdot \frac{\cos t}{-\sin t} = \frac{\cos t}{\ctg t} = \sin t.$$
Тождество доказано.
в)
$$\frac{\cos^2(\pi-t)+\sin^2\left(\frac{\pi}{2}-t\right)+\cos(\pi+t)\cos(2\pi-t)} {\tg^2\left(t-\frac{\pi}{2}\right)\ctg^2\left(\frac{3\pi}{2}+t\right)} = \frac{(-\cos t)^2+\cos^2 t-\cos t\cdot \cos t} {\left(-\tg\left(\frac{\pi}{2}-t\right)\right)^2\cdot \tg^2 t}$$
$$= \frac{\cos^2 t+\cos^2 t-\cos^2 t}{(-\ctg t)^2\cdot \tg^2 t} = \frac{\cos^2 t}{\ctg^2 t\cdot \tg^2 t} = \frac{\cos^2 t}{1} = \cos^2 t.$$
Тождество доказано.
г)
$$\frac{\sin^2\left(t-\frac{3\pi}{2}\right)\cos(2\pi-t)} {\tg^2\left(t-\frac{\pi}{2}\right)\cos^2\left(t-\frac{3\pi}{2}\right)} = \frac{\left(-\sin\left(\frac{3\pi}{2}-t\right)\right)^2\cos t} {\left(-\tg\left(\frac{\pi}{2}-t\right)\right)^2\cos^2\left(\frac{3\pi}{2}-t\right)}$$
$$= \frac{\cos^2 t\cdot \cos t}{(-\ctg t)^2\cdot (-\sin t)^2} = \frac{\cos^3 t}{\ctg^2 t\cdot \sin^2 t} = \frac{\cos^3 t}{\frac{\cos^2 t}{\sin^2 t}\cdot \sin^2 t} = \cos t.$$
Тождество доказано.
