cos frac{7 pi}{8}+cos frac{pi}{4}+cos left(fr | vedclass

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

Question

(C) આપેલ પદાવલિ: $\cos \frac{7 \pi}{8}+\cos \frac{\pi}{4}+\cos \left(\frac{-\pi}{8}ight)-1$. $\cos(-	heta) = \cos 	heta$ નો ઉપયોગ કરતા,પદાવલિ $\co

Vedclass · Accepted Answer

$4 \cos \frac{\pi}{16} \cos \frac{3 \pi}{8} \cos \frac{9 \pi}{16}$

Vedclass · Answer

(C) આપેલ પદાવલિ: $\cos \frac{7 \pi}{8}+\cos \frac{\pi}{4}+\cos \left(\frac{-\pi}{8}ight)-1$. $\cos(-	heta) = \cos 	heta$ નો ઉપયોગ કરતા,પદાવલિ $\cos \frac{7 \pi}{8} + \frac{1}{\sqrt{2}} + \cos \frac{\pi}{8} - 1$ બને છે. કારણ કે $\cos \frac{7 \pi}{8} = \cos(\pi - \frac{\pi}{8}) = -\cos \frac{\pi}{8}$,તેથી પદાવલિનું સાદું રૂપ: $-\cos \frac{\pi}{8} + \frac{1}{\sqrt{2}} + \cos \frac{\pi}{8} - 1 = \frac{1}{\sqrt{2}} - 1$. હવે,વિકલ્પ $(C)$ ચકાસતા: $4 \cos \frac{\pi}{16} \cos \frac{3 \pi}{8} \cos \frac{9 \pi}{16}$. $= 2 \left(2 \cos \frac{9 \pi}{16} \cos \frac{\pi}{16}ight) \cos \frac{3 \pi}{8}$. $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$ નો ઉપયોગ કરતા: $= 2 \left(\cos \frac{10 \pi}{16} + \cos \frac{8 \pi}{16}ight) \cos \frac{3 \pi}{8} = 2 \left(\cos \frac{5 \pi}{8} + \cos \frac{\pi}{2}ight) \cos \frac{3 \pi}{8}$. $\cos \frac{\pi}{2} = 0$ હોવાથી,આ $2 \cos \frac{5 \pi}{8} \cos \frac{3 \pi}{8}$ બને છે. $= \cos(\frac{5 \pi}{8} + \frac{3 \pi}{8}) + \cos(\frac{5 \pi}{8} - \frac{3 \pi}{8}) = \cos \pi + \cos \frac{\pi}{4} = -1 + \frac{1}{\sqrt{2}}$. આમ,વિકલ્પ $(C)$ સાચો છે.

$\cos \frac{7 \pi}{8}+\cos \frac{\pi}{4}+\cos \left(\frac{-\pi}{8}\right)-1=$

Similar Questions

$\frac{2\sin \theta \tan \theta (1 - \tan \theta ) + 2\sin \theta \sec^2 \theta}{(1 + \tan \theta)^2} = $

પદાવલિ $\cos^2 \phi + \cos^2(\theta + \phi) - 2 \cos \theta \cos \phi \cos(\theta + \phi)$ એ

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module