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

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

Question

(B) We know that $\cos(\pi - 	heta) = -\cos 	heta$,so $\cos^4(\pi - 	heta) = \cos^4 	heta$. Thus,$\cos^4 \frac{7\pi}{8} = \cos^4(\pi - \frac{\pi}{

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

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

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

Similar Questions

The value of $\sin 10^{\circ} \cdot \sin 30^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ}$ is

$\cos^3 \frac{\pi}{8} \cos \frac{3\pi}{8} + \sin^3 \frac{\pi}{8} \sin \frac{3\pi}{8} = $

If $\tan \theta = t,$ then $\tan 2\theta + \sec 2\theta = $

If $\frac{2 \sin \theta}{1+\cos \theta+\sin \theta}=y$,then $\frac{1-\cos \theta+\sin \theta}{1+\sin \theta}=$

Evaluate: $\sin(22 \frac{1}{2}^{\circ})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module