Упр.34 Задачи повышенной сложности ГДЗ Колмогоров 10-11 класс (Алгебра)

1. Вычислите: а) $$\frac{1}{\cos(290^\circ)}+\frac{1}{\sqrt{3}\sin(250^\circ)}$$; б) $$\sin^6(?) + \cos^6(?)$$, если $$\sin(?) + \cos(?) = m$$; в) $$\cos(84^\circ)\cos(24^\circ)\cos(48^\circ)\cos(12^\circ)$$; г) $$\cos^8(?) — \sin^8(?)$$, если $$\cos(2?) = m$$.

а)

$$\frac{1}{\cos 290^\circ}+\frac{1}{\sqrt{3}\sin 250^\circ} = \frac{1}{\cos(270^\circ+20^\circ)}+\frac{1}{\sqrt{3}\sin(270^\circ-20^\circ)}$$
$$= \frac{1}{\sin 20^\circ}-\frac{1}{\sqrt{3}\cos 20^\circ} = \frac{\sqrt{3}\cos 20^\circ-\sin 20^\circ}{\sqrt{3}\sin 20^\circ\cos 20^\circ}$$
$$= \frac{2\left(\frac{\sqrt{3}}{2}\cos 20^\circ-\frac{1}{2}\sin 20^\circ\right)}{\sqrt{3}\cdot \frac12 \sin 40^\circ} = \frac{4(\sin 60^\circ\cos 20^\circ-\cos 60^\circ\sin 20^\circ)}{\sqrt{3}\sin 40^\circ}$$
$$= \frac{4\sin(60^\circ-20^\circ)}{\sqrt{3}\sin 40^\circ} = \frac{4\sin 40^\circ}{\sqrt{3}\sin 40^\circ} = \frac{4}{\sqrt{3}}.$$

б)

$$x=\sin^6\alpha+\cos^6\alpha,\qquad \sin\alpha+\cos\alpha=m.$$
$$m^2=\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha, \qquad \sin 2\alpha=m^2-1.$$
$$x=(\sin^2\alpha+\cos^2\alpha)(\sin^4\alpha-\sin^2\alpha\cos^2\alpha+\cos^4\alpha)$$
$$=(\sin^4\alpha+2\sin^2\alpha\cos^2\alpha+\cos^4\alpha)-3\sin^2\alpha\cos^2\alpha$$
$$=(\sin^2\alpha+\cos^2\alpha)^2-\frac34\sin^2 2\alpha =1-\frac34(m^2-1)^2.$$

в)

$$\cos 84^\circ\cos 24^\circ\cos 48^\circ\cos 12^\circ = \frac{\sin 6^\circ\cos 6^\circ\cos 12^\circ\cos 24^\circ\cos 48^\circ}{\cos 6^\circ}$$
$$= \frac{\sin 12^\circ\cos 12^\circ\cos 24^\circ\cos 48^\circ}{2\cos 6^\circ} = \frac{\sin 24^\circ\cos 24^\circ\cos 48^\circ}{4\cos 6^\circ}$$
$$= \frac{\sin 48^\circ\cos 48^\circ}{8\cos 6^\circ} = \frac{\sin 96^\circ}{16\cos 6^\circ} = \frac{\sin(90^\circ+6^\circ)}{16\cos 6^\circ} = \frac{\cos 6^\circ}{16\cos 6^\circ} = \frac{1}{16}.$$

г)

$$x=\cos^8\alpha-\sin^8\alpha,\qquad \cos 2\alpha=m.$$
$$\cos 2\alpha=1-2\sin^2\alpha=2\cos^2\alpha-1=m,$$
$$\sin^2\alpha=\frac{1-m}{2},\qquad \cos^2\alpha=\frac{m+1}{2},\qquad \sin^2 2\alpha=1-m^2.$$
$$x=(\cos^2\alpha-\sin^2\alpha)(\cos^2\alpha+\sin^2\alpha)(\cos^4\alpha+\sin^4\alpha)$$
$$=m\left((\cos^2\alpha+\sin^2\alpha)^2-2\cos^2\alpha\sin^2\alpha\right) =m\left(1-\frac12\sin^2 2\alpha\right)$$
$$=m\left(1-\frac12(1-m^2)\right) =m\cdot \frac{1+m^2}{2} = \frac{m+m^3}{2}.$$

Ответ

а) $$\frac{4}{\sqrt{3}}$$; б) $$1-\frac34(m^2-1)^2$$; в) $$\frac{1}{16}$$; г) $$\frac{m+m^3}{2}$$.