Упр.26.7 ГДЗ Мерзляк 10 класс Углубленный уровень (Алгебра)

1) 1-cos(4?); 3) 1-cos(50°); 4) 1+sin(2?).
2) 1+cos(cos(6?)/sin(2?))+sin(6?)/cos(2?);

1. $$1-\cos 4\alpha=1-(1-2\sin^2 2\alpha)=2\sin^2 2\alpha.$$

2. $$1+\cos\left(\frac{\cos 6\alpha}{\sin 2\alpha}+\frac{\sin 6\alpha}{\cos 2\alpha}\right)$$
   $$=1+\cos\left(\frac{\cos 6\alpha\cos 2\alpha+\sin 6\alpha\sin 2\alpha}{\sin 2\alpha\cos 2\alpha}\right)$$
   $$=1+\cos\left(\frac{\cos(6\alpha-2\alpha)}{\sin 2\alpha\cos 2\alpha}\right)$$
   $$=1+\cos\left(\frac{\cos 4\alpha}{\frac12\sin 4\alpha}\right)=1+\cos(2\ctg 4\alpha).$$
   Тогда
   $$1+\cos(2\ctg 4\alpha)=1+\left(2\cos^2(\ctg 4\alpha)-1\right)=2\cos^2(\ctg 4\alpha).$$

3. $$1-\cos 50^\circ=1-(1-2\sin^2 25^\circ)=2\sin^2 25^\circ.$$

4. $$1+\sin 2\alpha=1+2\sin\alpha\cos\alpha$$
   $$=\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha$$
   $$=(\sin\alpha+\cos\alpha)^2.$$

Ответ

1) $$2\sin^2 2\alpha$$;
2) $$2\cos^2(\ctg 4\alpha)$$;
3) $$2\sin^2 25^\circ$$;
4) $$(\sin\alpha+\cos\alpha)^2$$.