Упр.23.24 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) cos^2(5?/4-2a)-cos^2(5?/4+2a)=sin(4a);
2) 1+2cos(2a)+cos(4a)=4cos^2(a)cos(2a);
3) (1+sin(2a)-cos(2a))/(1+sin(2a)+cos(2a))=tg(a);
4) (sin^2(2a)+4sin^2(a)-4)/(1-8sin^2(a)-cos(4a))=(1/2)ctg^4(a);
5) (cos(4a-?/2)sin(5?/2+2a))/((1+cos(2a))(1+cos(4a)))=tg(a);
6) (cos(4a)+1)/(ctg(a)-tg(a))=(1/2)sin(4a);
7) (2cos(2a)-sin(4a))/(2cos(2a)+sin(4a))=tg^2(45°-a).

1. $$\cos^2\left(\frac{5\pi}{4}-2a\right)-\cos^2\left(\frac{5\pi}{4}+2a\right)$$
   $$=\cos^2\left(\pi+\left(\frac{\pi}{4}-2a\right)\right)-\cos^2\left(\frac{3\pi}{2}-\left(\frac{\pi}{4}-2a\right)\right)$$
   $$=\cos^2\left(\frac{\pi}{4}-2a\right)-\sin^2\left(\frac{\pi}{4}-2a\right)$$
   $$=\cos\left(2\left(\frac{\pi}{4}-2a\right)\right)=\cos\left(\frac{\pi}{2}-4a\right)=\sin 4a.$$

2. $$1+2\cos 2a+\cos 4a=1+2\cos 2a+\left(2\cos^2 2a-1\right)$$
   $$=2\cos 2a+2\cos^2 2a=2\cos 2a\,(1+\cos 2a)$$
   $$=2\cos 2a\,(1+2\cos^2 a-1)=4\cos^2 a\cos 2a.$$

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

4. $$\frac{\sin^2 2a+4\sin^2 a-4}{1-8\sin^2 a-\cos 4a}$$
   $$=\frac{(2\sin a\cos a)^2+4\sin^2 a-4}{1-8\sin^2 a-(1-2\sin^2 2a)}$$
   $$=\frac{4\sin^2 a\cos^2 a+4\sin^2 a-4}{-8\sin^2 a+8\sin^2 a\cos^2 a}$$
   $$=\frac{4\cos^2 a(\sin^2 a-1)}{8\sin^2 a(\cos^2 a-1)}$$
   $$=\frac{-4\cos^4 a}{-8\sin^2 a\cos^2 a}=\frac{1}{2}\ctg^4 a.$$

5. $$\frac{\cos\left(4a-\frac{\pi}{2}\right)\sin\left(\frac{5\pi}{2}+2a\right)}{(1+\cos 2a)(1+\cos 4a)}$$
   $$=\frac{\sin 4a\cdot \cos 2a}{(1+\cos 2a)(1+\cos 4a)}$$
   $$=\frac{2\sin 2a\cos 2a\cdot \cos 2a}{(1+2\cos^2 a-1)(1+2\cos^2 2a-1)}$$
   $$=\frac{2\sin 2a\cos^2 2a}{2\cos^2 a\cdot 2\cos^2 2a}$$
   $$=\frac{\sin 2a}{2\cos^2 a}=\frac{2\sin a\cos a}{2\cos^2 a}=\tg a.$$

6. $$\frac{\cos 4a+1}{\ctg a-\tg a}$$
   $$=\frac{2\cos^2 2a}{\frac{\cos a}{\sin a}-\frac{\sin a}{\cos a}}$$
   $$=\frac{2\cos^2 2a}{\frac{\cos^2 a-\sin^2 a}{\sin a\cos a}}$$
   $$=\frac{2\cos^2 2a\cdot \sin a\cos a}{\cos 2a}$$
   $$=2\sin a\cos a\cos 2a=\sin 2a\cos 2a=\frac{1}{2}\sin 4a.$$

7. $$\frac{2\cos 2a-\sin 4a}{2\cos 2a+\sin 4a}$$
   $$=\frac{2\cos 2a-2\sin 2a\cos 2a}{2\cos 2a+2\sin 2a\cos 2a}$$
   $$=\frac{2\cos 2a(1-\sin 2a)}{2\cos 2a(1+\sin 2a)}=\frac{1-\sin 2a}{1+\sin 2a}$$
   $$=\frac{(\cos a-\sin a)^2}{(\cos a+\sin a)^2}$$
   $$=\tg^2\left(\frac{\pi}{4}-a\right)=\tg^2(45^\circ-a).$$

Ответ

Все тождества доказаны.