What is \H(t-a)e^{-bt}cos^2(t-a) ?

The solution to \H(t-a)e^{-bt}cos^2(t-a) is (\H(t-a)cos^2(t-a))/(e^{bt)}

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simplify \H(t-a)e^{-bt}cos^2(t-a)

simplify

H (t - a) e^{- b t} cos^{2} (t - a)

\frac{H ( t - a ) cos ^{2} ( t - a )}{e ^{b t}}

Solution steps

H (t - a) e^{- b t} cos^{2} (t - a)

Apply exponent rule:

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

= \frac{1}{e ^{b t}} H (t - a) cos^{2} (t - a)

Multiply fractions:

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

= \frac{1 \cdot H ( t - a ) cos ^{2} ( t - a )}{e ^{b t}}

Multiply:

1 \cdot H (t - a) = H (t - a)

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

5.3 x 1 0^{- 6} \cdot (- 975)

(3 x^{2}) \cdot (4 x)

x (2 x^{3})

(e^{j (2 t + \frac{π}{4})})^{2}

What is \H(t-a)e^{-bt}cos^2(t-a) ?
The solution to \H(t-a)e^{-bt}cos^2(t-a) is (\H(t-a)cos^2(t-a))/(e^{bt)}