化简 (e^{iθ}-1)(e^{-iθ}-1)

解答

化简

(e^{i θ} - 1) (e^{- i θ} - 1)

(cos (θ) - 1)^{2} + sin^{2} (θ)

求解步骤

(e^{i θ} - 1) (e^{- i θ} - 1)

使用虚数运算法则:

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

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

e^{- i θ} = cos (θ) - i sin (θ)

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

使用复数算数法则:

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

a = cos (θ) - 1, b = sin (θ)

= (cos (θ) - 1)^{2} + sin^{2} (θ)

s im pl i f y 2 (x + \frac{1}{x}) (1 - \frac{1}{x ^{2}})

e x p an d lo g_{10} (x^{3} y^{- 12})

s im pl i f y (x - 1) (x + \frac{1}{2})

e x p an d e^{t} (3 cos (5 t) - 2 sin (5 s))

e x p an d \frac{10 s}{( s + 2 ) ^{3} ( s + 8 )}