What is e^{-i*(pi/2-x)} ?

The solution to e^{-i*(pi/2-x)} is cos((-pi+2x)/2)+isin((-pi+2x)/2)

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simplify e^{-i*(pi/2-x)}

simplify

e^{- i \cdot (\frac{π}{2} - x)}

cos (\frac{- π + 2 x}{2}) + i sin (\frac{- π + 2 x}{2})

Solution steps

e^{- i (\frac{π}{2} - x)}

Apply imaginary number rule:

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

= cos (\frac{2 x - π}{2}) + i sin (\frac{2 x - π}{2})

(1 - sin (θ))^{2} + (cos (θ))^{2}

(s - \frac{5}{3} + 2 k) (s + \frac{5}{3} + j)

(\frac{d y}{d t} + \frac{2}{t} y = t^{2} - t + 1) t^{2}

(2 - \frac{1}{3} x)^{4}

A (- 7, 3) B (\frac{1}{2}, - \frac{3}{4})

What is e^{-i*(pi/2-x)} ?
The solution to e^{-i*(pi/2-x)} is cos((-pi+2x)/2)+isin((-pi+2x)/2)