Упр.26.2 ГДЗ Мерзляк 10 класс Углубленный уровень (Алгебра)

1) cos(6?)+2sin^2(3?); 5) sin(4?)/(cos^4(?)-sin^4(?));
2) cos(70°)/(cos(35°)+sin(35°)); 6) sin(?/4-?)cos(?/4-?);
3) (1+sin(2?))/(sin(?)+cos(?))^2; 7) sin^2(?-45°)-cos^2(?-45°);
4) sin(?)cos(?)cos(2?); 8) cos^2(?/4-3?/2)-sin^2(?/4-3?/2).

1. $$\cos 6\alpha+2\sin^2 3\alpha=1-2\sin^2 3\alpha+2\sin^2 3\alpha=1.$$
2. $$\frac{\cos 70^\circ}{\cos 35^\circ+\sin 35^\circ} =\frac{\cos^2 35^\circ-\sin^2 35^\circ}{\cos 35^\circ+\sin 35^\circ} =\cos 35^\circ-\sin 35^\circ.$$
3. $$\frac{1+\sin 2\alpha}{(\sin \alpha+\cos \alpha)^2} =\frac{1+2\sin \alpha\cos \alpha}{\sin^2 \alpha+\cos^2 \alpha+2\sin \alpha\cos \alpha} =1.$$
4. $$\sin \alpha\cos \alpha\cos 2\alpha =\frac12\sin 2\alpha\cos 2\alpha =\frac14\sin 4\alpha.$$
5. $$\frac{\sin 4\alpha}{\cos^4 \alpha-\sin^4 \alpha} =\frac{2\sin 2\alpha\cos 2\alpha}{(\cos^2 \alpha-\sin^2 \alpha)(\cos^2 \alpha+\sin^2 \alpha)} =\frac{2\sin 2\alpha\cos 2\alpha}{\cos 2\alpha} =2\sin 2\alpha.$$
6. $$\sin\left(\frac{\pi}{4}-\alpha\right)\cos\left(\frac{\pi}{4}-\alpha\right) =\frac12\sin\left(\frac{\pi}{2}-2\alpha\right) =\frac12\cos 2\alpha.$$
7. $$\sin^2(\beta-45^\circ)-\cos^2(\beta-45^\circ) =-\cos\bigl(2(\beta-45^\circ)\bigr) =-\cos(2\beta-90^\circ) =-\sin 2\beta.$$
8. $$\cos^2\left(\frac{\pi}{4}-\frac{3\alpha}{2}\right)-\sin^2\left(\frac{\pi}{4}-\frac{3\alpha}{2}\right) =\cos\left(\frac{\pi}{2}-3\alpha\right) =\sin 3\alpha.$$

Ответ

1) $$1$$; 2) $$\cos 35^\circ-\sin 35^\circ$$; 3) $$1$$; 4) $$\frac14\sin 4\alpha$$; 5) $$2\sin 2\alpha$$; 6) $$\frac12\cos 2\alpha$$; 7) $$-\sin 2\beta$$; 8) $$\sin 3\alpha$$.