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:

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

Answer 2

Let's calculate the expression step by step:

1. Calculate $1 + 2\sin(\pi/4)$: $1 + 2\sin(\pi/4) = 1 + 2\cdot\frac{1}{\sqrt{2}} = 1 + \sqrt{2}.$

2. Calculate $1 - 2\cos(\pi/4)$: $1 - 2\cos(\pi/4) = 1 - 2\cdot\frac{1}{\sqrt{2}} = 1 - \sqrt{2}.$

3. Square both results: $(1 + \sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2},$ and $(1 - \sqrt{2})^2 = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}.$

4. Subtract the second result from the first result: $(3 + 2\sqrt{2}) - (3 - 2\sqrt{2}) = 3 + 2\sqrt{2} - 3 + 2\sqrt{2} = 4\sqrt{2}.$

So, the final result is $4\sqrt{2}$.