What is [1-cos(αt)]e^{-βt} ?

The solution to [1-cos(αt)]e^{-βt} is (-cos(αt)+1)/(e^{βt)}

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simplify [1-cos(αt)]e^{-βt}

simplify

[1 - cos (α t)] e^{- βt}

\frac{- cos ( α t ) + 1}{e ^{βt}}

Solution steps

[1 - cos (α t)] e^{- βt}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{βt}} [- cos (α t) + 1]

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot ( 1 - cos ( α t ) )}{e ^{βt}}

1 \cdot (1 - cos (α t)) = 1 - cos (α t)

= \frac{1 - cos ( α t )}{e ^{βt}}

(e^{5})^{x}

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

(1.8 \times 1 0^{6}) \times (2.5 \times 1 0^{8})

5 e^{- 4^{2}}

1.49 \cdot e^{- \frac{97100}{8.314} \cdot (\frac{1}{829} - \frac{1}{900})}

What is [1-cos(αt)]e^{-βt} ?
The solution to [1-cos(αt)]e^{-βt} is (-cos(αt)+1)/(e^{βt)}