Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 168161 by daus last updated on 05/Apr/22

Commented by daus last updated on 05/Apr/22

help show me the way. thanks

$${help}\:{show}\:{me}\:{the}\:{way}.\:{thanks} \\ $$

Answered by alephzero last updated on 05/Apr/22

x^2 −2x cos θ+1 = 0  ⇒ x = ((2 cos θ±(√(4 cos^2  θ−4)))/2) =  = ((2 cos θ±(√(4(cos^2  θ−1))))/2) =  = cos θ±(√(cos^2  θ−1)) =  = cos θ±i ∣sin θ∣

$${x}^{\mathrm{2}} −\mathrm{2}{x}\:\mathrm{cos}\:\theta+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{x}\:=\:\frac{\mathrm{2}\:\mathrm{cos}\:\theta\pm\sqrt{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{4}}}{\mathrm{2}}\:= \\ $$$$=\:\frac{\mathrm{2}\:\mathrm{cos}\:\theta\pm\sqrt{\mathrm{4}\left(\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{1}\right)}}{\mathrm{2}}\:= \\ $$$$=\:\mathrm{cos}\:\theta\pm\sqrt{\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{1}}\:= \\ $$$$=\:\mathrm{cos}\:\theta\pm{i}\:\mid\mathrm{sin}\:\theta\mid \\ $$

Commented by peter frank last updated on 05/Apr/22

good

$$\mathrm{good} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com