Упр.27.21 ГДЗ Мордковича 10 класс профильный уровень (Алгебра)

а) $$\frac{1+\cos 40^\circ+\cos 80^\circ}{\sin 80^\circ+\sin 40^\circ}\cdot\operatorname{tg}40^\circ$$; б) $$\frac{1-\cos 25^\circ+\cos 50^\circ}{\sin 50^\circ+\sin 25^\circ}-\operatorname{tg}65^\circ$$.

а) $$\frac{1+\cos 40^\circ+\cos 80^\circ}{\sin 80^\circ+\sin 40^\circ}\cdot \tg 40^\circ$$

Используем формулы $$\cos 80^\circ=\cos^2 40^\circ-\sin^2 40^\circ$$ и $$\sin 80^\circ=2\sin 40^\circ\cos 40^\circ$$:

$$\frac{1+\cos 40^\circ+\cos^2 40^\circ-\sin^2 40^\circ}{2\sin 40^\circ\cos 40^\circ+\sin 40^\circ}\cdot \tg 40^\circ= \frac{(\cos^2 40^\circ+\sin^2 40^\circ)+\cos 40^\circ+(\cos^2 40^\circ-\sin^2 40^\circ)}{\sin 40^\circ(2\cos 40^\circ+1)}\cdot \tg 40^\circ$$

$$=\frac{2\cos^2 40^\circ+\cos 40^\circ}{\sin 40^\circ(2\cos 40^\circ+1)}\cdot \tg 40^\circ =\frac{\cos 40^\circ(2\cos 40^\circ+1)}{\sin 40^\circ(2\cos 40^\circ+1)}\cdot \tg 40^\circ =\frac{\cos 40^\circ}{\sin 40^\circ}\cdot \tg 40^\circ=1$$

б) $$\frac{1-\cos 25^\circ+\cos 50^\circ}{\sin 50^\circ-\sin 25^\circ}-\tg 65^\circ$$

Так как $$\cos 50^\circ=\cos^2 25^\circ-\sin^2 25^\circ$$ и $$\sin 50^\circ=2\sin 25^\circ\cos 25^\circ$$, получаем:

$$\frac{1-\cos 25^\circ+\cos^2 25^\circ-\sin^2 25^\circ}{2\sin 25^\circ\cos 25^\circ-\sin 25^\circ}-\tg 65^\circ$$

$$=\frac{(\cos^2 25^\circ+\sin^2 25^\circ)-\cos 25^\circ+(\cos^2 25^\circ-\sin^2 25^\circ)}{\sin 25^\circ(2\cos 25^\circ-1)}-\tg 65^\circ$$

$$=\frac{2\cos^2 25^\circ-\cos 25^\circ}{\sin 25^\circ(2\cos 25^\circ-1)}-\tg 65^\circ =\frac{\cos 25^\circ(2\cos 25^\circ-1)}{\sin 25^\circ(2\cos 25^\circ-1)}-\tg 65^\circ =\frac{\cos 25^\circ}{\sin 25^\circ}-\tg 65^\circ$$

$$=\ctg 25^\circ-\tg 65^\circ=\tg 65^\circ-\tg 65^\circ=0$$

Ответ: а) $$1$$; б) $$0$$.