Question

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Answer 1

$\frac{\sqrt{2}sin\frac{\pi }{8}-2cos\frac{5\pi}{8}}{sin\frac{5\pi}{8}}=$

$\frac{\sqrt{2}sin\frac{\pi }{8}-2cos\left( \frac{\pi}{2}+\frac{\pi }{8} \right) }{sin\left( \frac{\pi}{2}+\frac{\pi }{8} \right)}=$

$\frac{\sqrt{2}sin\frac{\pi }{8}+2sin\frac{\pi }{8}}{cos\frac{\pi}{8}}=$

$\frac{sin\frac{\pi }{8}(\sqrt2+2)}{cos\frac{\pi}{8}}=$

$\frac{2cos\frac{\pi}{8} \cdot sin\frac{\pi }{8} (\sqrt2+2)}{ 2cos\frac{\pi}{8} \cdot cos\frac{\pi}{8}}=$

$\frac{sin\frac{\pi }{4} (\sqrt2+2)}{ 2cos^2\frac{\pi}{8}}=$

$\frac{ \frac{\sqrt2}{2} (\sqrt2+2)}{ 2cos^2\frac{\pi}{8}-1+1}=$

$\frac{\frac{\sqrt2}{2} (\sqrt2+2)}{cos\frac{\pi}{4}+1}=$

$\frac{\frac{\sqrt2}{2} (\sqrt2+2)}{\frac{\sqrt2}{2} +1}=$

$\frac{\frac{\sqrt2}{2} (\sqrt2+2)}{\frac{\sqrt2+2}{2}}=$

$\frac{\frac{\sqrt2}{2}}{\frac{1}{2}}=\sqrt2$