sin ^4 frac{pi}{8} + sin ^4 frac{3pi}{8} + si | vedclass

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

Question

(C) We know that $\sin \frac{5\pi}{8} = \sin(\pi - \frac{3\pi}{8}) = \sin \frac{3\pi}{8}$ and $\sin \frac{7\pi}{8} = \sin(\pi - \frac{\pi}{8}) = \sin

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) We know that $\sin \frac{5\pi}{8} = \sin(\pi - \frac{3\pi}{8}) = \sin \frac{3\pi}{8}$ and $\sin \frac{7\pi}{8} = \sin(\pi - \frac{\pi}{8}) = \sin \frac{\pi}{8}$.Thus,the expression becomes $2(\sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8})$.Using the identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we have $\sin^4 	heta = (\frac{1 - \cos 2	heta}{2})^2 = \frac{1}{4}(1 - 2\cos 2	heta + \cos^2 2	heta) = \frac{1}{4}(1 - 2\cos 2	heta + \frac{1 + \cos 4	heta}{2}) = \frac{1}{8}(3 - 4\cos 2	heta + \cos 4	heta)$.For $	heta = \frac{\pi}{8}$: $\sin^4 \frac{\pi}{8} = \frac{1}{8}(3 - 4\cos \frac{\pi}{4} + \cos \frac{\pi}{2}) = \frac{1}{8}(3 - 4(\frac{1}{\sqrt{2}}) + 0) = \frac{3 - 2\sqrt{2}}{8}$.For $	heta = \frac{3\pi}{8}$: $\sin^4 \frac{3\pi}{8} = \frac{1}{8}(3 - 4\cos \frac{3\pi}{4} + \cos \frac{3\pi}{2}) = \frac{1}{8}(3 - 4(-\frac{1}{\sqrt{2}}) + 0) = \frac{3 + 2\sqrt{2}}{8}$.Summing these: $2(\frac{3 - 2\sqrt{2}}{8} + \frac{3 + 2\sqrt{2}}{8}) = 2(\frac{6}{8}) = 2(\frac{3}{4}) = \frac{3}{2}$.

$\sin ^4 \frac{\pi}{8} + \sin ^4 \frac{3\pi}{8} + \sin ^4 \frac{5\pi}{8} + \sin ^4 \frac{7\pi}{8} = $

Similar Questions

If $\left| a\sin^2 \theta + b\sin \theta \cos \theta + c\cos^2 \theta - \frac{1}{2}(a + c) \right| \le \frac{1}{2}k,$ then $k^2$ is equal to

If angle $\theta$ is divided into two parts such that the tangent of one part is $k$ times the tangent of the other and $\phi$ is their difference,then $\sin \theta = $

Which of the following number$(s)$ is/are rational?

$\frac{\cot \theta - \operatorname{cosec} \theta + 1}{\cot \theta + \operatorname{cosec} \theta - 1}$ is equal to

If the maximum value of $y = \frac{7 + 6 \tan x - \tan^2 x}{1 + \tan^2 x}$ is $\lambda$,then the value of $\log_{\sqrt{2}}(\lambda)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module