Упр.537 ГДЗ Алимов 10-11 класс (Алгебра)

1. Упростить выражение: 1) $$\sin\left(\frac{\pi}{3}+a\right)+\sin\left(\frac{\pi}{3}-a\right)$$; 2) $$\cos\left(\frac{\pi}{4}-b\right)-\cos\left(\frac{\pi}{4}+b\right)$$; 3) $$\sin^2\left(\frac{\pi}{4}+a\right)-\sin^2\left(\frac{\pi}{4}-a\right)$$; 4) $$\cos^2\left(a-\frac{\pi}{4}\right)-\cos^2\left(a+\frac{\pi}{4}\right)$$.

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

$$=2\sin\frac{\pi}{3}\cos a=2\cdot\frac{\sqrt{3}}{2}\cos a=\sqrt{3}\cos a.$$

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

$$=-2\sin\frac{\pi}{4}\sin(-b)=2\sin\frac{\pi}{4}\sin b=2\cdot\frac{\sqrt{2}}{2}\sin b=\sqrt{2}\sin b.$$

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

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

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

$$=2\cos\frac{\pi}{4}\sin a\cdot 2\sin\frac{\pi}{4}\cos a=2\sin a\cos a=\sin 2a.$$

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

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

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

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

Ответ

1) $$\sqrt{3}\cos a$$; 2) $$\sqrt{2}\sin b$$; 3) $$\sin 2a$$; 4) $$\sin 2a$$.