Упр.13 ГДЗ Колмогоров 10-11 класс (Алгебра)

а) $$8\sin\frac{\pi}{6}\cos\frac{2\pi}{3}\tg\frac{4\pi}{3}\ctg\frac{7\pi}{4}$$

Используем формулы приведения и значения тригонометрических функций:

$$8\sin\frac{\pi}{6}\cos\frac{2\pi}{3}\tg\frac{4\pi}{3}\ctg\frac{7\pi}{4}= 8\sin\frac{\pi}{6}\cos\left(\pi-\frac{\pi}{3}\right)\tg\left(\pi+\frac{\pi}{3}\right)\ctg\left(2\pi-\frac{\pi}{4}\right)$$
$$=8\sin\frac{\pi}{6}\left(-\cos\frac{\pi}{3}\right)\tg\frac{\pi}{3}\ctg\left(-\frac{\pi}{4}\right)$$
$$=8\cdot\frac12\cdot\left(-\frac12\right)\cdot\sqrt{3}\cdot(-1)=2\sqrt{3}.$$

б) $$10\ctg\frac{3\pi}{4}\sin\frac{5\pi}{4}\cos\frac{7\pi}{4}$$

$$10\ctg\frac{3\pi}{4}\sin\frac{5\pi}{4}\cos\frac{7\pi}{4}= 10\ctg\left(\pi-\frac{\pi}{4}\right)\sin\left(\pi+\frac{\pi}{4}\right)\cos\left(2\pi-\frac{\pi}{4}\right)$$
$$=10\cdot\ctg\frac{\pi}{4}\cdot\left(-\sin\frac{\pi}{4}\right)\cdot\cos\frac{\pi}{4}$$
$$=10\cdot 1\cdot\left(-\frac{\sqrt{2}}{2}\right)\cdot\frac{\sqrt{2}}{2}=-5.$$

в) $$\frac{\sin^2(\pi-t)}{1+\sin\left(\frac{3\pi}{2}+t\right)}-\cos(2\pi-t)$$

$$\sin(\pi-t)=\sin t,\qquad \sin\left(\frac{3\pi}{2}+t\right)=-\cos t,\qquad \cos(2\pi-t)=\cos t.$$

Тогда
$$\frac{\sin^2(\pi-t)}{1+\sin\left(\frac{3\pi}{2}+t\right)}-\cos(2\pi-t) =\frac{\sin^2 t}{1-\cos t}-\cos t.$$

Так как $$\sin^2 t=1-\cos^2 t=(1-\cos t)(1+\cos t),$$ то
$$\frac{\sin^2 t}{1-\cos t}-\cos t=1+\cos t-\cos t=1.$$

Ответ: а) $$2\sqrt{3}$$; б) $$-5$$; в) $$1$$.