簡約 (1-e^{ix})(1-e^{-ix})

解

簡約

(1 - e^{i x}) (1 - e^{- i x})

(1 - cos (x))^{2} + sin^{2} (x)

解答ステップ

(1 - e^{i x}) (1 - e^{- i x})

虚数の規則を適用する:

e^{ia} = cos (a) + i sin (a)

= (- (cos (x) + i sin (x)) + 1) (- e^{- i x} + 1)

e^{- i x} = cos (x) - i sin (x)

= (- (cos (x) + i sin (x)) + 1) (- (cos (x) - i sin (x)) + 1)

複素数の算術規則を適用する:

(a + bi) (a - bi) = a^{2} + b^{2}

a = 1 - cos (x), b = - sin (x)

= (1 - cos (x))^{2} + (- sin (x))^{2}

簡素化

(1 - cos (x))^{2} + (- sin (x))^{2} : (1 - cos (x))^{2} + sin^{2} (x)

= (1 - cos (x))^{2} + sin^{2} (x)