What is 2(1+cos(x))e^{-ix} ?

The solution to 2(1+cos(x))e^{-ix} is (2cos(x)+2cos^2(x))+i(-2sin(x)-2cos(x)sin(x))

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simplify 2(1+cos(x))e^{-ix}

simplify

2 (1 + cos (x)) e^{- i x}

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

Solution steps

2 (1 + cos (x)) e^{- i x}

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

= 2 (cos (x) + 1) (cos (x) - i sin (x))

Rewrite

2 (1 + cos (x)) (cos (x) - i sin (x))

in standard complex form:

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

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

ln (x^{7} + 8)

\frac{4 a}{( 2 a + 1 ) ( 2 a - 1 )}

(8 x^{2} - 2 x y + 3) \frac{d x}{d y}

- (- 2 x^{2} + 3 x - 7 = 0)

14 a + 18 a

What is 2(1+cos(x))e^{-ix} ?
The solution to 2(1+cos(x))e^{-ix} is (2cos(x)+2cos^2(x))+i(-2sin(x)-2cos(x)sin(x))