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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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}{\

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(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

Similar Questions

If $n = 1983!$,then the value of the expression $\frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \frac{1}{\log_4 n} + \dots + \frac{1}{\log_{1983} n}$ is:

If ${x^{\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}} = \sqrt{3}}$,then $x$ is:

The set of real values of $x$ that satisfy the inequality ${\log _{1/2}}({x^2} - 6x + 12) \ge - 2$ is:

The value of $\log _4 18$ is

$\tanh^{-1}(\frac{1}{2}) + \operatorname{coth}^{-1}(3) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module