What is e^{((-4t))}[sin(2t)-cos(2t)]^2 ?

The solution to e^{((-4t))}[sin(2t)-cos(2t)]^2 is ((sin(2t)-cos(2t))^2)/(e^{4t)}

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simplify e^{((-4t))}[sin(2t)-cos(2t)]^2

simplify

e^{((- 4 t))} [sin (2 t) - cos (2 t)]^{2}

\frac{( sin ( 2 t ) - cos ( 2 t ) ) ^{2}}{e ^{4 t}}

Solution steps

e^{((- 4 t))} [sin (2 t) - cos (2 t)]^{2}

Apply exponent rule:

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

= \frac{1}{e ^{4 t}} [sin (2 t) - cos (2 t)]^{2}

Multiply fractions:

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

= \frac{1 \cdot ( sin ( 2 t ) - cos ( 2 t ) ) ^{2}}{e ^{4 t}}

Multiply:

1 \cdot (sin (2 t) - cos (2 t))^{2} = (sin (2 t) - cos (2 t))^{2}

= \frac{( sin ( 2 t ) - cos ( 2 t ) ) ^{2}}{e ^{4 t}}

(6.02 \cdot 1 0^{23}) (4.1 \cdot 1 0^{- 16})

2 e^{x} \cdot e^{2 x}

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

(1 - 1)^{2} e^{- 1}

2.74 \cdot 1 0^{- 3}

What is e^{((-4t))}[sin(2t)-cos(2t)]^2 ?
The solution to e^{((-4t))}[sin(2t)-cos(2t)]^2 is ((sin(2t)-cos(2t))^2)/(e^{4t)}