Question

Поможіть, будь ласка!!!!

Answer 1

sin(π/4) = (√2)/2 = cos(π/4)

$\displaystyle \sqrt{\bigg(1+2\cos \dfrac{\pi}4\bigg) ^2} -\sqrt{\bigg(1-2\sin \dfrac{\pi}4\bigg) ^2} =\\\\=\bigg| 1+2\cos \dfrac{\pi}4\bigg| -\bigg| 1-2\sin \dfrac{\pi}4\bigg| =\bigg| 1+\sqrt2\bigg| -\bigg| 1-\sqrt2\bigg| =\\\\=1+\sqrt2+(1-\sqrt2)=2$

Ответ: 2.

Комментарий, второй модуль раскрылся со знаком минус т.к. 1<√2.

Answer 2

$\sqrt{(1+2Cos\frac{\pi }{4})^{2}}-\sqrt{(1-2Sin\frac{\pi }{4})^{2}} =\sqrt{(1+2*\frac{\sqrt{2} }{2})^{2}}-\sqrt{(1-2*\frac{\sqrt{2}}{2})^{2}}=\sqrt{(1+\sqrt{2})^{2}}-\sqrt{(1-\sqrt{2})^{2}}=|1+\sqrt{2}|-|1-\sqrt{2}|=1+\sqrt{2}-(\sqrt{2}-1)=1+\sqrt{2}-\sqrt{2}+1=2\\\\Otvet:\boxed{2}$