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Question Number 128805 by bramlexs22 last updated on 10/Jan/21
Prove that e^(iθ) +1 = e^(−((iθ)/2)) + e^((iθ)/2) ?
$$\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{e}^{\mathrm{i}\theta} +\mathrm{1}\:=\:\mathrm{e}^{−\frac{\mathrm{i}\theta}{\mathrm{2}}} \:+\:\mathrm{e}^{\frac{\mathrm{i}\theta}{\mathrm{2}}} \:? \\ $$
Commented by mr W last updated on 10/Jan/21
LHS=(1+cos θ)+i sin θ RHS=2 cos (θ/2) LHS≠RHS !
$${LHS}=\left(\mathrm{1}+\mathrm{cos}\:\theta\right)+{i}\:\mathrm{sin}\:\theta \\ $$$${RHS}=\mathrm{2}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}} \\ $$$${LHS}\neq{RHS}\:! \\ $$
Answered by benjo_mathlover last updated on 10/Jan/21
e^(iθ) +1 = cos θ+i sin θ +1 e^(−((iθ)/2)) = cos (θ/2)−i sin (θ/2) e^((iθ)/2) = cos (θ/2) +i sin (θ/2) e^(−((iθ)/2)) + e^((iθ)/2) = 2cos (θ/2) ≠ 1+cos θ+i sin θ
$$\:\mathrm{e}^{{i}\theta} +\mathrm{1}\:=\:\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\:+\mathrm{1} \\ $$$$\:\mathrm{e}^{−\frac{{i}\theta}{\mathrm{2}}} \:=\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}−{i}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}} \\ $$$$\:\mathrm{e}^{\frac{{i}\theta}{\mathrm{2}}} \:=\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}}\:+{i}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\: \\ $$$$\:\mathrm{e}^{−\frac{{i}\theta}{\mathrm{2}}} \:+\:\mathrm{e}^{\frac{{i}\theta}{\mathrm{2}}} \:=\:\mathrm{2cos}\:\frac{\theta}{\mathrm{2}}\:\neq\:\mathrm{1}+\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta \\ $$

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