Question

Verificati : $cos$$\frac{\pi }{8} = \frac{1}{2} \sqrt{2+\sqrt{2}}$ , $sin$$\frac{\pi }{2} = \frac{1}{2} \sqrt{2-{\sqrt{2}}$

Answer 1

$1)\,\,\,\,\cos\frac{\pi}{8} = \dfrac{1}{2}\sqrt{2+\sqrt 2}\Big|^2\\ \\ \Leftrightarrow\, \cos^2\frac{\pi}{8} = \dfrac{1}{4}\left(2+\sqrt 2\right)\Big|\cdot 2\\ \\ \Leftrightarrow\,2\cos^2 \frac{\pi}{8} = \dfrac{(2+\sqrt{2})}{2}\Big|-1\\ \\ \Leftrightarrow\, 2\cos^2\frac{\pi}{8}-1= 1+\sqrt{2}-1\\ \\ \Leftrightarrow \,\cos(2\cdot \frac{\pi}{8}) = \dfrac{\sqrt{2}}{2}\\ \\ \Leftrightarrow \, \cos\frac{\pi}{4} = \dfrac{\sqrt 2}{2}\quad (A)$

$2)\,\,\,\,\sin \frac{\pi}{8} = \dfrac{1}{2}\sqrt{2-\sqrt{2}}\\ \\ \boxed{\Big|\sin \frac{x}{2}\Big|=\sqrt{\dfrac{1-\cos x}{2}}}\\ \\ \sin\left(\dfrac{\dfrac{\pi}{4}}{2}\right) = \sqrt{\dfrac{1-\cos \frac{\pi}{4}}{2}} = \sqrt{\dfrac{1-\dfrac{\sqrt 2}{2}}{4}}=\dfrac{1}{2}\sqrt{2-\sqrt{2}}\quad (A)$