left(frac{1+cos frac{pi}{8}-i sin frac{pi}{8} | vedclass

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

Question

(B) ધારો કે $z = \frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}$.અડધા ખૂણાના સૂત્રો $1+\cos 	heta = 2 \c

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) ધારો કે $z = \frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}$.અડધા ખૂણાના સૂત્રો $1+\cos 	heta = 2 \cos^2 \frac{	heta}{2}$ અને $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ નો ઉપયોગ કરતા:$z = \frac{2 \cos^2 \frac{\pi}{16} - 2i \sin \frac{\pi}{16} \cos \frac{\pi}{16}}{2 \cos^2 \frac{\pi}{16} + 2i \sin \frac{\pi}{16} \cos \frac{\pi}{16}}$$z = \frac{2 \cos \frac{\pi}{16} (\cos \frac{\pi}{16} - i \sin \frac{\pi}{16})}{2 \cos \frac{\pi}{16} (\cos \frac{\pi}{16} + i \sin \frac{\pi}{16})}$$z = \frac{\cos \frac{\pi}{16} - i \sin \frac{\pi}{16}}{\cos \frac{\pi}{16} + i \sin \frac{\pi}{16}} = \frac{e^{-i \frac{\pi}{16}}}{e^{i \frac{\pi}{16}}} = e^{-i \frac{2\pi}{16}} = e^{-i \frac{\pi}{8}}$.હવે,$z^8 = (e^{-i \frac{\pi}{8}})^8 = e^{-i \pi}$.ઓઈલરના સૂત્ર $e^{i 	heta} = \cos 	heta + i \sin 	heta$ નો ઉપયોગ કરતા:$e^{-i \pi} = \cos(-\pi) + i \sin(-\pi) = -1 + 0 = -1$.

$\left(\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^8$ ની કિંમત શોધો.

Similar Questions

${\left( \frac{-1 + i\sqrt{3}}{2} \right)^{20}} + {\left( \frac{-1 - i\sqrt{3}}{2} \right)^{20}} = $

$(1 + 2\omega + \omega^2)^{3n} - (1 + \omega + 2\omega^2)^{3n}$ ની કિંમત શોધો.

$\sum_{k=1}^6 \left[ \sin \frac{2 k \pi}{7} - i \cos \frac{2 k \pi}{7} \right]$ ની કિંમત શોધો.

જો $\omega$ એ એકમનું સંકર ઘનમૂળ હોય,તો $\omega^{\left(\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\ldots \infty\right)}+\omega^{\left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\ldots \infty\right)}$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module