What is (e^{(-2+i)t}+e^{(-2-i)t})/2 ?

The solution to (e^{(-2+i)t}+e^{(-2-i)t})/2 is e^{-2t}cos(t)

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simplify (e^{(-2+i)t}+e^{(-2-i)t})/2

simplify

\frac{e ^{(- 2 + i) t} + e ^{(- 2 - i) t}}{2}

e^{- 2 t} cos (t)

Solution steps

\frac{e ^{(- 2 + i) t} + e ^{(- 2 - i) t}}{2}

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

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

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

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

Factor

e^{- 2 t} (cos (t) + sin (t) i) + e^{- 2 t} (cos (t) - i sin (t)) : 2 e^{- 2 t} cos (t)

= \frac{2 e ^{- 2 t} cos ( t )}{2}

Divide the numbers:

\frac{2}{2} = 1

= e^{- 2 t} cos (t)

\frac{( 2 + s ^{5} - s ^{4} + 2 s ^{3} )}{s ^{4} ( s ^{2} - s + 2 )}

\frac{x ( 4 - x )}{2}

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

(1 + cos^{2} (x)) (tan (x))

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

What is (e^{(-2+i)t}+e^{(-2-i)t})/2 ?
The solution to (e^{(-2+i)t}+e^{(-2-i)t})/2 is e^{-2t}cos(t)