What is (-e^{ik}(ik-1)+e^{-ik}(-ik-1))/(k^2) ?

The solution to (-e^{ik}(ik-1)+e^{-ik}(-ik-1))/(k^2) is i(2sin(k)-2kcos(k))/(k^2)

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simplify (-e^{ik}(ik-1)+e^{-ik}(-ik-1))/(k^2)

simplify

\frac{- e ^{ik} ( ik - 1 ) + e ^{- ik} ( - ik - 1 )}{k ^{2}}

i \frac{2 sin ( k ) - 2 k cos ( k )}{k ^{2}}

Solution steps

\frac{- e ^{ik} ( ik - 1 ) + e ^{- ik} ( - ik - 1 )}{k ^{2}}

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

= \frac{- ( cos ( k ) + i sin ( k ) ) ( ik - 1 ) + ( cos ( - k ) + i sin ( - k ) ) ( - ik - 1 )}{k ^{2}}

(cos (k) + sin (k) i) (ik - 1) = - cos (k) - k sin (k) + i (k cos (k) - sin (k))

(cos (- k) + sin (- k) i) (- ik - 1) = - k sin (k) - cos (k) + i (sin (k) - k cos (k))

= \frac{- ( i ( k cos ( k ) - sin ( k ) ) - k sin ( k ) - cos ( k ) ) + ( i ( sin ( k ) - k cos ( k ) ) - k sin ( k ) - cos ( k ) )}{k ^{2}}

Remove parentheses:

(- a) = - a

= \frac{- ( - cos ( k ) - k sin ( k ) + ( k cos ( k ) - sin ( k ) ) i ) - k sin ( k ) - cos ( k ) + ( sin ( k ) - k cos ( k ) ) i}{k ^{2}}

Expand

- (- cos (k) - k sin (k) + (k cos (k) - sin (k)) i) - k sin (k) - cos (k) + (sin (k) - k cos (k)) i : - 2 ik cos (k) + 2 i sin (k)

= \frac{- 2 ik cos ( k ) + 2 i sin ( k )}{k ^{2}}

Rewrite

\frac{- 2 ik cos ( k ) + 2 i sin ( k )}{k ^{2}}

in standard complex form:

\frac{- 2 k cos ( k ) + 2 sin ( k )}{k ^{2}} i

= \frac{- 2 k cos ( k ) + 2 sin ( k )}{k ^{2}} i

(t - 2)^{3} e^{3 t}

(\frac{1}{x} + 2)^{2}

(y + 1) (x^{2} + ln (x))

2 (x + y^{2}) d x + (y - x) d y

What is (-e^{ik}(ik-1)+e^{-ik}(-ik-1))/(k^2) ?
The solution to (-e^{ik}(ik-1)+e^{-ik}(-ik-1))/(k^2) is i(2sin(k)-2kcos(k))/(k^2)