log _{frac{1}{8}csc^2 frac{pi}{8}} sin^2 frac | vedclass

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

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(C) Let $E = \log_{\frac{1}{8}\csc^2 \frac{\pi}{8}} \sin^2 \frac{3\pi}{8}$.First,simplify 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 $\frac{1}{8}(4+2\sqrt{2}) = \frac{2+\sqrt{2}}{4} = \frac{\sqrt{2}(\sqrt{2}+1)}{4} = \frac{\sqrt{2}+1}{2\sqrt{2}}$.Next,simplify the argument: $\sin^2 \frac{3\pi}{8} = \frac{1-\cos(3\pi/4)}{2} = \frac{1-(-\frac{1}{\sqrt{2}})}{2} = \frac{1+\frac{1}{\sqrt{2}}}{2} = \frac{\sqrt{2}+1}{2\sqrt{2}}$.Therefore,$E = \log_{\frac{\sqrt{2}+1}{2\sqrt{2}}} \left( \frac{\sqrt{2}+1}{2\sqrt{2}} \right) = 1$.

$\log _{\frac{1}{8}\csc^2 \frac{\pi}{8}} \sin^2 \frac{3\pi}{8}$ equals to

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