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Question Number 204117 by mokys last updated on 06/Feb/24
solve for θ in terms of y when y = e^θ (cosθ+sinθ)
$${solve}\:{for}\:\theta\:{in}\:{terms}\:{of}\:{y}\:{when}\:{y}\:=\:{e}^{\theta} \left({cos}\theta+{sin}\theta\right)\:\:\: \\ $$
Commented by mr W last updated on 06/Feb/24
impossible!
$${impossible}! \\ $$
Commented by mokys last updated on 06/Feb/24
thank you alot sir
$${thank}\:{you}\:{alot}\:{sir} \\ $$
Commented by mokys last updated on 06/Feb/24
sory sir i want θ for y
$${sory}\:{sir}\:{i}\:{want}\:\theta\:{for}\:{y} \\ $$
Commented by Frix last updated on 07/Feb/24
Maybe y=e^θ (cos θ +i sin θ), because then  it′s possible:  e^(iθ) =cos θ +i sin θ =e^(iθ)   ⇒  y=e^((1+i)θ)   θ=((ln y)/2)+nπ−(((ln y)/2)−nπ)i
$$\mathrm{Maybe}\:{y}=\mathrm{e}^{\theta} \left(\mathrm{cos}\:\theta\:+\mathrm{i}\:\mathrm{sin}\:\theta\right),\:\mathrm{because}\:\mathrm{then} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{possible}: \\ $$$$\mathrm{e}^{\mathrm{i}\theta} =\mathrm{cos}\:\theta\:+\mathrm{i}\:\mathrm{sin}\:\theta\:=\mathrm{e}^{\mathrm{i}\theta} \\ $$$$\Rightarrow \\ $$$${y}=\mathrm{e}^{\left(\mathrm{1}+\mathrm{i}\right)\theta} \\ $$$$\theta=\frac{\mathrm{ln}\:{y}}{\mathrm{2}}+{n}\pi−\left(\frac{\mathrm{ln}\:{y}}{\mathrm{2}}−{n}\pi\right)\mathrm{i} \\ $$

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