簡約 |e^{i(ωt+θ)}|^2

解

簡約

e^{i (ω t + θ)}^{2}

cos^{2} (ω t + θ) + sin^{2} (ω t + θ)

解答ステップ

e^{i (ω t + θ)}^{2}

e^{i (ω t + θ)} = cos (ω t + θ) + i sin (ω t + θ)

= ∣ cos (ω t + θ) + sin (ω t + θ) i ∣^{2}

簡素化

∣ cos (ω t + θ) + sin (ω t + θ) i ∣ : cos^{2} (ω t + θ) + sin^{2} (ω t + θ)

= (cos^{2} (ω t + θ) + sin^{2} (ω t + θ))^{2}

累乗根の規則を適用する:

a = a^{\frac{1}{2}}

= ((cos^{2} (ω t + θ) + sin^{2} (ω t + θ))^{\frac{1}{2}})^{2}

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

= (cos^{2} (ω t + θ) + sin^{2} (ω t + θ))^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

= cos^{2} (ω t + θ) + sin^{2} (ω t + θ)