Упр.519 ГДЗ Алимов 10-11 класс (Алгебра)

519. 1) 2cos2(пи/4-a/2) = 1+sina;
2) 2sin2(пи/4 — a/2) = 1-sina;
3) (3-4cos2a + cos4a)/(3+4cos2a+cos4a) = tg4a;
4) (1+sin2a+co2a)/(1+sin2a — cos2a) = ctga.

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

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

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

   По формулам
   $$1-\cos 2a=2\sin^2 a,\qquad 1+\cos 2a=2\cos^2 a,$$
   получаем
   $$\left(\frac{1-\cos 2a}{1+\cos 2a}\right)^2 =\left(\frac{2\sin^2 a}{2\cos^2 a}\right)^2 =\left(\frac{\sin a}{\cos a}\right)^4 =\tg^4 a.$$

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

Ответ

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

2) $$2\sin^2\left(\frac{\pi}{4}-\frac{a}{2}\right)=1-\sin a$$

3) $$\frac{3-4\cos 2a+\cos 4a}{3+4\cos 2a+\cos 4a}=\tg^4 a$$

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