If cos ^4 frac{pi}{8}+cos ^4 frac{3 pi}{8}+co | vedclass

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

Question

(A) Given expression: $\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8} = k$. Using $\cos(\frac{\pi}{2} +

Vedclass · Accepted Answer

$\frac{2 \pi}{3}$

Vedclass · Answer

(A) Given expression: $\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8} = k$. Using $\cos(\frac{\pi}{2} + 	heta) = -\sin 	heta$,we have $\cos^4(\frac{\pi}{2} + 	heta) = \sin^4 	heta$. So,$\cos^4 \frac{5 \pi}{8} = \cos^4(\frac{\pi}{2} + \frac{\pi}{8}) = \sin^4 \frac{\pi}{8}$ and $\cos^4 \frac{7 \pi}{8} = \cos^4(\frac{\pi}{2} + \frac{3 \pi}{8}) = \sin^4 \frac{3 \pi}{8}$. Substituting these,$k = (\cos^4 \frac{\pi}{8} + \sin^4 \frac{\pi}{8}) + (\cos^4 \frac{3 \pi}{8} + \sin^4 \frac{3 \pi}{8})$. Using $a^2 + b^2 = (a+b)^2 - 2ab$,we get $k = [(\cos^2 \frac{\pi}{8} + \sin^2 \frac{\pi}{8})^2 - 2 \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}] + [(\cos^2 \frac{3 \pi}{8} + \sin^2 \frac{3 \pi}{8})^2 - 2 \sin^2 \frac{3 \pi}{8} \cos^2 \frac{3 \pi}{8}]$. Since $\sin^2 	heta \cos^2 	heta = \frac{1}{4} \sin^2(2 	heta)$,$k = [1 - \frac{1}{2} \sin^2 \frac{\pi}{4}] + [1 - \frac{1}{2} \sin^2 \frac{3 \pi}{4}]$. $k = 2 - \frac{1}{2}(\frac{1}{2}) - \frac{1}{2}(\frac{1}{2}) = 2 - \frac{1}{4} - \frac{1}{4} = 2 - \frac{1}{2} = \frac{3}{2}$. Now,$\sin^{-1}(\sqrt{\frac{k}{2}}) + \cos^{-1}(\frac{k}{3}) = \sin^{-1}(\sqrt{\frac{3/2}{2}}) + \cos^{-1}(\frac{3/2}{3}) = \sin^{-1}(\frac{\sqrt{3}}{2}) + \cos^{-1}(\frac{1}{2})$. $= \frac{\pi}{3} + \frac{\pi}{3} = \frac{2 \pi}{3}$.

If $\cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8}+\cos ^4 \frac{7 \pi}{8}=k$ then $\sin ^{-1}\left(\sqrt{\frac{k}{2}}\right)+\cos ^{-1}\left(\frac{k}{3}\right)=$

Similar Questions

If the equation $\sin^{-1}(x - 1) + \cos^{-1}(x - 3) + \tan^{-1}\left(\frac{x}{-x^2 + 2}\right) = m$ holds,then the value of $m$ is

If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$,then $x(y+z)+y(z+x)+z(x+y)$ equals to

If $\alpha$ and $\beta$ are the roots of the quadratic equation $3x^2 - 16x + 5 = 0$,then $\tan^{-1} \alpha + \tan^{-1} \beta - \tan^{-1}\left(\frac{\alpha + \beta}{1 - \alpha \beta}\right) = $

The numerical value of $\tan \left(2 \tan ^{-1}\left(\frac{1}{5}\right)+\frac{\pi}{4}\right)$ is:

If $x = \sin \left( 2 \tan^{-1} 2 \right)$ and $y = \sin \left( \frac{1}{2} \tan^{-1} \frac{4}{3} \right)$,then -

Vedclass Test Series

Exam Paper Generator

Online Exam Module