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Question Number 177280 by mokys last updated on 02/Oct/22
by ussing demover find (cos(π/2) + i cos(π/8))^(−5)
$${by}\:{ussing}\:{demover}\:{find}\: \\ $$$$ \\ $$$$\left({cos}\frac{\pi}{\mathrm{2}}\:\:+\:{i}\:{cos}\frac{\pi}{\mathrm{8}}\right)^{−\mathrm{5}} \:\: \\ $$
Commented by Frix last updated on 03/Oct/22
cos (π/2) =0 cos (π/8) =((√(2+(√2)))/2) ⇒ answer is 8(4−3(√2))(√(2−(√2)))i
$$\mathrm{cos}\:\frac{\pi}{\mathrm{2}}\:=\mathrm{0} \\ $$$$\mathrm{cos}\:\frac{\pi}{\mathrm{8}}\:=\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{8}\left(\mathrm{4}−\mathrm{3}\sqrt{\mathrm{2}}\right)\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\mathrm{i} \\ $$
Answered by Ar Brandon last updated on 03/Oct/22
(cos(π/2)+icos(π/8))^(−5) =(isin((3π)/8))^(−5) =(1/(isin^5 (((3π)/8)))) cos(((3π)/4))=1−2sin^2 (((3π)/8)) ⇒−((√2)/2)=1−2sin^2 (((3π)/8)) ⇒sin(((3π)/8))=±((√(2+(√2)))/2)=((√(2+(√2)))/2) (sinϑ>0: 0<ϑ<π) (cos(π/2)+icos(π/8))^(−5) =−(2^5 /( (√((2+(√2))^5 ))))i
$$\left(\mathrm{cos}\frac{\pi}{\mathrm{2}}+{i}\mathrm{cos}\frac{\pi}{\mathrm{8}}\right)^{−\mathrm{5}} =\left({i}\mathrm{sin}\frac{\mathrm{3}\pi}{\mathrm{8}}\right)^{−\mathrm{5}} =\frac{\mathrm{1}}{{i}\mathrm{sin}^{\mathrm{5}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)} \\ $$$$\mathrm{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\:\Rightarrow−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\ $$$$\Rightarrow\mathrm{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)=\pm\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\:\left(\mathrm{sin}\vartheta>\mathrm{0}:\:\mathrm{0}<\vartheta<\pi\right) \\ $$$$\left(\mathrm{cos}\frac{\pi}{\mathrm{2}}+{i}\mathrm{cos}\frac{\pi}{\mathrm{8}}\right)^{−\mathrm{5}} =−\frac{\mathrm{2}^{\mathrm{5}} }{\:\sqrt{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)^{\mathrm{5}} }}{i} \\ $$
Commented by puissant last updated on 04/Oct/22
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Answered by mr W last updated on 03/Oct/22
(cos (π/2)+i cos (π/8))^(−5) =(0+((√(2+(√2)))/2)i)^(−5) =(((√(2+(√2)))/2))^(−5) (i)^(−5) =(2^5 /(((√(2+(√2))))^5 ))×(i/i^6 ) =((32)/(((√(2+(√2))))^5 ))×(i/((−1)^3 )) =−((32i)/(((√(2+(√2))))^5 ))
$$\left(\mathrm{cos}\:\frac{\pi}{\mathrm{2}}+{i}\:\mathrm{cos}\:\frac{\pi}{\mathrm{8}}\right)^{−\mathrm{5}} \\ $$$$=\left(\mathrm{0}+\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}{i}\right)^{−\mathrm{5}} \\ $$$$=\left(\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\right)^{−\mathrm{5}} \left({i}\right)^{−\mathrm{5}} \\ $$$$=\frac{\mathrm{2}^{\mathrm{5}} }{\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}\right)^{\mathrm{5}} }×\frac{{i}}{{i}^{\mathrm{6}} } \\ $$$$=\frac{\mathrm{32}}{\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}\right)^{\mathrm{5}} }×\frac{{i}}{\left(−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=−\frac{\mathrm{32}{i}}{\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}\right)^{\mathrm{5}} } \\ $$
Answered by ranjankumar last updated on 03/Oct/22

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