Вычислите. 1-sin^6(22,5*)+cos^6(22,5*)

Решите задачу:

$1-sin^6\, 22,5 +cos^6\, 22,5=1-(sin^2\, 22,5)^3+(cos^2\, 22,5)^3=\\\\\\\star \; \; sin^2 \alpha = \frac{1-cos2 \alpha }{2}\; \; ,\; \; cos^2 \alpha = \frac{1+cos2 \alpha }{2}\; ,\; cos45^\circ = \frac{\sqrt2}{2}=\frac{1}{\sqrt2} \; \star$

$=1-\Big ( \frac{1-cos45}{2}\Big )^3+\Big (\frac{1+cos45}{2}\Big )^3=1- \frac{(1-\frac{1}{\sqrt2})^3}{8}+\frac{(1+\frac{1}{\sqrt2})^3}{8} =\\\\=1- \frac{(\sqrt2-1)^3}{2\sqrt2\cdot 8} + \frac{(\sqrt2+1)^3}{2\sqrt2\cdot 8} =1- \frac{2\sqrt2-3\cdot 2+3\sqrt2-1}{16\sqrt2}+\frac{2\sqrt2+3\cdot 2+3\sqrt2+1}{16\sqrt2}=$

$=1+\frac{-2\sqrt2+6-3\sqrt2+1+2\sqrt2+6+3\sqrt2+1}{16\sqrt2} =1+\frac{14}{16\sqrt2} =1+\frac{7}{8\sqrt2} =\\\\= \frac{8\sqrt2+7}{8\sqrt2}= \frac{16+7\sqrt2}{16}$