Упр.22.9 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) (sin(?+a)cos(2?-a))/(tg(?-a)cos(?-a));
2) sin(?-?)cos(?-?/2)-sin(?/2+?)cos(?-?);
3) sin(90°+a)sin(180°-a)(tg(180°+a)+tg(270°-a));
4) sin^2(?-a)+tg^2(?-a)tg^2(3?/2+x)+sin(?/2+a)cos(x-2?);
5) (sin(?/2-x)+sin(?-x))^2+(cos(3?/2-x)+cos(2?-x))^2;
6) (tg(?-x)sin(3?/2+x))/(cos(?+x)ctg(3?/2+x)).

1. $$\frac{\sin(\pi+a)\cos(2\pi-a)}{\tg(\pi-a)\cos(\pi-a)}$$
   $$=\frac{(-\sin a)\cos a}{(-\tg a)(-\cos a)}$$
   $$=\frac{-\sin a\cos a}{-\frac{\sin a}{\cos a}\cdot(-\cos a)}=-\cos a.$$

2. $$\sin(\pi-\beta)\cos\left(\beta-\frac{\pi}{2}\right)-\sin\left(\frac{\pi}{2}+\beta\right)\cos(\pi-\beta)$$
   $$=\sin\beta\cdot\sin\beta-\cos\beta\cdot(-\cos\beta)$$
   $$=\sin^2\beta+\cos^2\beta=1.$$

3. $$\sin(90^\circ+a)\sin(180^\circ-a)\bigl(\tg(180^\circ+a)+\tg(270^\circ-a)\bigr)$$
   $$=\cos a\cdot\sin a\left(\tg a+\ctg a\right)$$
   $$=\cos a\cdot\sin a\left(\frac{\sin a}{\cos a}+\frac{\cos a}{\sin a}\right)$$
   $$=\sin^2 a+\cos^2 a=1.$$

4. $$\sin^2(\pi-a)+\tg^2(\pi-a)\tg^2\left(\frac{3\pi}{2}+x\right)+\sin\left(\frac{\pi}{2}+x\right)\cos(x-2\pi)$$
   $$=\sin^2 x+\tg^2 x\cdot\ctg^2 x+\cos x\cdot\cos x$$
   $$=\sin^2 x+1+\cos^2 x=2.$$

5. $$\left(\sin\left(\frac{\pi}{2}-x\right)+\sin(\pi-x)\right)^2+\left(\cos\left(\frac{3\pi}{2}-x\right)+\cos(2\pi-x)\right)^2$$
   $$=(\cos x+\sin x)^2+(-\sin x+\cos x)^2$$
   $$=(\cos^2 x+2\sin x\cos x+\sin^2 x)+(\cos^2 x-2\sin x\cos x+\sin^2 x)$$
   $$=2(\sin^2 x+\cos^2 x)=2.$$

6. $$\frac{\tg(\pi-x)\sin\left(\frac{3\pi}{2}+x\right)}{\cos(\pi+x)\ctg\left(\frac{3\pi}{2}+x\right)}$$
   $$=\frac{(-\tg x)(-\cos x)}{(-\cos x)(-\tg x)}=1.$$

Ответ

1) $$-\cos a$$; 2) $$1$$; 3) $$1$$; 4) $$2$$; 5) $$2$$; 6) $$1$$.