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}}

e^{- \frac{0}{0}}

e^{- 10 x} e^{- 10 x}

900 e^{(- 2.4849 \cdot 1)}

A t^{(2)} e^{(- 2 t)}

8 x^{2} (4 x + 1)^{2} (2 x + 1)^{2}

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)}