What is e^{i(kr-ωt)} ?

The solution to e^{i(kr-ωt)} is cos(kr-ωt)+isin(kr-ωt)

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simplify e^{i(kr-ωt)}

simplify

e^{i (k r - ω t)}

cos (k r - ω t) + i sin (k r - ω t)

Solution steps

e^{i (k r - ω t)}

Apply imaginary number rule:

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

= cos (k r - ω t) + i sin (k r - ω t)

19 (e - 1)

(n - 2)^{(n - 2)^{2}}

\frac{e ^{y}}{( y + 5 ) ^{7}}

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

\frac{5 x + 1}{( x - 2 ) ^{2}}