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Question Number 177280 by mokys last updated on 02/Oct/22

$$byussingdemoverfind$$$$$$$${(cos\frac{\pi }{2}+icos\frac{\pi }{8})}^{-5}$$

Commented by Frix last updated on 03/Oct/22

$$\cos \frac{\pi }{2}=0$$$$\cos \frac{\pi }{8}=\frac{\sqrt{2+\sqrt{2}}}{2}$$$$\Rightarrow \mathrm{answer}\mathrm{is}8(4-3\sqrt{2})\sqrt{2-\sqrt{2}}\mathrm{i}$$

Answered by Ar Brandon last updated on 03/Oct/22

$${(\cos \frac{\pi }{2}+i\cos \frac{\pi }{8})}^{-5}={\left(i\sin \frac{3\pi }{8}\right)}^{-5}=\frac{1}{i{\sin }^5\left(\frac{3\pi }{8}\right)}$$$$\cos \left(\frac{3\pi }{4}\right)=1-{2\sin }^2\left(\frac{3\pi }{8}\right)\Rightarrow -\frac{\sqrt{2}}{2}=1-{2\sin }^2\left(\frac{3\pi }{8}\right)$$$$\Rightarrow \sin \left(\frac{3\pi }{8}\right)=\pm \frac{\sqrt{2+\sqrt{2}}}{2}=\frac{\sqrt{2+\sqrt{2}}}{2}(\sin \vartheta >0:0<\vartheta <\pi )$$$${(\cos \frac{\pi }{2}+i\cos \frac{\pi }{8})}^{-5}=-\frac{2^5}{\sqrt{{(2+\sqrt{2})}^5}}i$$

Commented by puissant last updated on 04/Oct/22

$$.$$

Answered by mr W last updated on 03/Oct/22

$${(\cos \frac{\pi }{2}+i\cos \frac{\pi }{8})}^{-5}$$$$={(0+\frac{\sqrt{2+\sqrt{2}}}{2}i)}^{-5}$$$$={\left(\frac{\sqrt{2+\sqrt{2}}}{2}\right)}^{-5}{\left(i\right)}^{-5}$$$$=\frac{2^5}{{\left(\sqrt{2+\sqrt{2}}\right)}^5}\times \frac{i}{i^6}$$$$=\frac{32}{{\left(\sqrt{2+\sqrt{2}}\right)}^5}\times \frac{i}{{(-1)}^3}$$$$=-\frac{32i}{{\left(\sqrt{2+\sqrt{2}}\right)}^5}$$

Answered by ranjankumar last updated on 03/Oct/22

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