What is ae^{-at(cos(bt+θ))} ?

The solution to ae^{-at(cos(bt+θ))} is a/(e^{atcos(bt+θ))}

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simplify ae^{-at(cos(bt+θ))}

simplify

a e^{- a t (c o s (b t + θ))}

\frac{a}{e ^{a t c o s (b t + θ)}}

Solution steps

a e^{- a t (c o s (b t + θ))}

Apply exponent rule:

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

= a \frac{1}{e ^{a t c o s (b t + θ)}}

Multiply fractions:

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

= \frac{1 \cdot a}{e ^{a t c o s (b t + θ)}}

Multiply:

1 \cdot a = a

= \frac{a}{e ^{a t c o s (b t + θ)}}

l (l + 1) - m (m + 1) l (l + 1) - m (m - 1)

(3.25 \times 1 0^{7}) (8.16 \times 1 0^{9})

5 t e^{- 2 t}

e^{- 0.03 t}

What is ae^{-at(cos(bt+θ))} ?
The solution to ae^{-at(cos(bt+θ))} is a/(e^{atcos(bt+θ))}