The value of log_{frac{1}{8}csc^2frac{pi}{8}} | vedclass

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(C) Let the expression be $E = \log_{\frac{1}{8}\csc^2\frac{\pi}{8}} \sin^2\frac{3\pi}{8}$.First,calculate the base: $\csc^2\frac{\pi}{8} = \frac{1}{\

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(C) Let the expression be $E = \log_{\frac{1}{8}\csc^2\frac{\pi}{8}} \sin^2\frac{3\pi}{8}$.First,calculate the base: $\csc^2\frac{\pi}{8} = \frac{1}{\sin^2\frac{\pi}{8}} = \frac{1}{\frac{1-\cos(\pi/4)}{2}} = \frac{2}{1-\frac{1}{\sqrt{2}}} = \frac{2\sqrt{2}}{\sqrt{2}-1} = 2\sqrt{2}(\sqrt{2}+1) = 4+2\sqrt{2}$.So,the base is $\frac{1}{8}(4+2\sqrt{2}) = \frac{2+\sqrt{2}}{4}$.Next,calculate the argument: $\sin^2\frac{3\pi}{8} = \frac{1-\cos(3\pi/4)}{2} = \frac{1-(-\frac{1}{\sqrt{2}})}{2} = \frac{1+\frac{1}{\sqrt{2}}}{2} = \frac{\sqrt{2}+1}{2\sqrt{2}} = \frac{\sqrt{2}+1}{2\sqrt{2}} 	imes \frac{\sqrt{2}}{\sqrt{2}} = \frac{2+\sqrt{2}}{4}$.Thus,$E = \log_{\frac{2+\sqrt{2}}{4}} \left(\frac{2+\sqrt{2}}{4}ight) = 1$.

The value of $\log_{\frac{1}{8}\csc^2\frac{\pi}{8}} \sin^2\frac{3\pi}{8}$ is equal to

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