Question

(1+2sin π/4)^2-√(1-2cos π/4)^2

Answer 1

Ответ:

$4+\sqrt{2}$

Решение:

$(1+2sin\pi/4)^2-\sqrt{(1-2cos\pi/4)^2}=(1+2*\frac{\sqrt{2}}{2})^2-|1-2*\frac{\sqrt{2}}{2}|=\\\\=(1+\sqrt{2})^2-|1-\sqrt{2}|=1+2\sqrt{2}+2-(\sqrt{2}-1)=3+2\sqrt{2}-\sqrt{2}+1=4+\sqrt{2}$

Объяснение:

$(a+b)^2=a^2+2ab+b^2\\\\sin\frac{\pi}{4}=cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\\\\\sqrt{x^2}=|x| \\\\\sqrt{2}\approx1,41= > (1-\sqrt{2}) < 0 = > |1-\sqrt{2}|=-(1-\sqrt{2})=\sqrt{2}-1$