Упр.23.4 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) sin(80°)/cos(40°); 6) sin(a)cos(a)(cos^2(a)-sin^2(a));
2) cos(4?)+sin^2(2?); 7) sin(4a)/(cos^4(a)-sin^4(a));
3) cos(6a)+2sin^2(3a); 8) sin(?/4-a)cos(?/4-a);
4) cos(70°)/(cos(35°)+sin(35°)); 9) sin^2(?-45°)-cos^2(?-45°);
5) (1+sin(2a))/(sin(a)+cos(a))^2; 10) (2tg(1,5a))/(1+tg^2(1,5a)).

1. $$\frac{\sin 80^\circ}{\cos 40^\circ}=\frac{\sin(2\cdot 40^\circ)}{\cos 40^\circ}=\frac{2\sin 40^\circ\cos 40^\circ}{\cos 40^\circ}=2\sin 40^\circ.$$
2. $$\cos 4\beta+\sin^2 2\beta=(\cos^2 2\beta-\sin^2 2\beta)+\sin^2 2\beta=\cos^2 2\beta.$$
3. $$\cos 6a+2\sin^2 3a=(\cos^2 3a-\sin^2 3a)+2\sin^2 3a=\cos^2 3a+\sin^2 3a=1.$$
4. $$\frac{\cos 70^\circ}{\cos 35^\circ+\sin 35^\circ}=\frac{\cos(2\cdot 35^\circ)}{\cos 35^\circ+\sin 35^\circ}=\frac{\cos^2 35^\circ-\sin^2 35^\circ}{\cos 35^\circ+\sin 35^\circ}$$
   $$=\frac{(\cos 35^\circ-\sin 35^\circ)(\cos 35^\circ+\sin 35^\circ)}{\cos 35^\circ+\sin 35^\circ}=\cos 35^\circ-\sin 35^\circ.$$
5. $$\frac{1+\sin 2a}{(\sin a+\cos a)^2}=\frac{1+2\sin a\cos a}{\sin^2 a+\cos^2 a+2\sin a\cos a}=\frac{1+\sin 2a}{1+\sin 2a}=1.$$
6. $$\sin a\cos a(\cos^2 a-\sin^2 a)=\frac12\sin 2a\cdot \cos 2a=\frac14\sin 4a.$$
7. $$\frac{\sin 4a}{\cos^4 a-\sin^4 a}=\frac{\sin 4a}{(\cos^2 a-\sin^2 a)(\cos^2 a+\sin^2 a)}=\frac{\sin 4a}{\cos 2a}$$
   $$=\frac{2\sin 2a\cos 2a}{\cos 2a}=2\sin 2a.$$
8. $$\sin\left(\frac{\pi}{4}-a\right)\cos\left(\frac{\pi}{4}-a\right)=\frac12\sin\left(2\left(\frac{\pi}{4}-a\right)\right)=\frac12\sin\left(\frac{\pi}{2}-2a\right)=\frac12\cos 2a.$$
9. $$\sin^2(\beta-45^\circ)-\cos^2(\beta-45^\circ)=-\cos\bigl(2(\beta-45^\circ)\bigr)$$
   $$=-\cos(2\beta-90^\circ)=-\sin 2\beta.$$
10. $$\frac{2\tg 1{,}5a}{1+\tg^2 1{,}5a}=\sin(2\cdot 1{,}5a)=\sin 3a.$$

Ответ

1. $$2\sin 40^\circ$$
2. $$\cos^2 2\beta$$
3. $$1$$
4. $$\cos 35^\circ-\sin 35^\circ$$
5. $$1$$
6. $$\frac14\sin 4a$$
7. $$2\sin 2a$$
8. $$\frac12\cos 2a$$
9. $$-\sin 2\beta$$
10. $$\sin 3a$$