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Question 1

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$sin\frac{\pi}{12}*(cos^{6}\frac{\pi}{24}-sin^{6}\frac{\pi}{24})=$

Question 2

=(14+sqrt(3))/64................................................

Question 3

$sin\frac{\pi}{12}*(cos^{6}\frac{\pi}{24}-sin^{6}\frac{\pi}{24}) =\\\\ sin\frac{\pi}{12}*((cos^{3}\frac{\pi}{24})^2-(sin^{3}\frac{\pi}{24})^2) =\\\\ sin\frac{\pi}{12}*(cos^{3}\frac{\pi}{24}-sin^{3}\frac{\pi}{24})(cos^{3}\frac{\pi}{24}+sin^{3}\frac{\pi}{24}) =$

$sin\frac{\pi}{12}*(cos\frac{\pi}{24}-sin\frac{\pi}{24})(cos^2\frac{\pi}{24} + cos\frac{\pi}{24}sin\frac{\pi}{24} + sin^2\frac{\pi}{24})*\\\\ (cos\frac{\pi}{24}+sin\frac{\pi}{24})(cos^2\frac{\pi}{24} - cos\frac{\pi}{24}sin\frac{\pi}{24} + sin^2\frac{\pi}{24}) =\\\\ sin\frac{\pi}{12}*(cos\frac{\pi}{24}-sin\frac{\pi}{24})(1 + cos\frac{\pi}{24}sin\frac{\pi}{24})*\\\\ (cos\frac{\pi}{24}+sin\frac{\pi}{24})(1 - cos\frac{\pi}{24}sin\frac{\pi}{24}) =$

$sin\frac{\pi}{12}(cos^2\frac{\pi}{24}-sin^2\frac{\pi}{24})(1 + cos\frac{\pi}{24}sin\frac{\pi}{24}) (1 - cos\frac{\pi}{24}sin\frac{\pi}{24}) = \\\\ sin\frac{\pi}{12}(cos^2\frac{\pi}{24}-sin^2\frac{\pi}{24})(1 - cos^2\frac{\pi}{24}sin^2 \frac{\pi}{24}) =\\\\$

$[cos2x = cos^2x - sin^2x]\\\\ sin\frac{\pi}{12}cos\frac{\pi}{12}(1 - cos^2\frac{\pi}{24}sin^2\frac{\pi}{24}) = \\\\ \frac{1}{2}sin\frac{\pi}{6}(1 - \frac{1}{4}(sin^2\frac{\pi}{12})) =$

$\frac{1}{2}sin\frac{\pi}{6}(1 - \frac{1}{4}(\frac{1 - cos\frac{\pi}{6}}{2})) = \frac{1}{2}*\frac{1}{2}*(1 - \frac{1}{8}(1 - \frac{\sqrt{3}}{2})) =\\\\ \frac{1}{4}(1 - \frac{1}{8} + \frac{\sqrt{3}}{16})= \frac{1}{4}*(\frac{14+\sqrt{3}}{16}) = \boxed{\frac{14 + \sqrt{3}}{64}}$