What is expand (-e^tsin(t)+cos(t)e^t)^2 ?

The solution to expand (-e^tsin(t)+cos(t)e^t)^2 is e^{2t}(-sin(2t)+1)

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expand (-e^tsin(t)+cos(t)e^t)^2

expand

(- e^{t} sin (t) + cos (t) e^{t})^{2}

e^{2 t} (- sin (2 t) + 1)

Solution steps

(- e^{t} sin (t) + cos (t) e^{t})^{2}

= (- e^{t} sin (t) + e^{t} cos (t))^{2}

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = - e^{t} sin (t), b = cos (t) e^{t}

= (- e^{t} sin (t))^{2} + 2 (- e^{t} sin (t)) cos (t) e^{t} + (cos (t) e^{t})^{2}

Simplify

(- e^{t} sin (t))^{2} + 2 (- e^{t} sin (t)) cos (t) e^{t} + (cos (t) e^{t})^{2} : e^{2 t} sin^{2} (t) - 2 e^{2 t} sin (t) cos (t) + e^{2 t} cos^{2} (t)

= e^{2 t} sin^{2} (t) - 2 e^{2 t} sin (t) cos (t) + e^{2 t} cos^{2} (t)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

= cos^{2} (t) e^{2 t} + sin^{2} (t) e^{2 t} - sin (2 t) e^{2 t}

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

= e^{2 t} (- sin (2 t) + 1)

\frac{2 s ^{2} + 1}{s ( s + 1 ) ^{2}}

6 (5 x^{2} + 6 x + 6)^{3}

(2 cos (t) - t sin (t))^{2}

\frac{4}{16} x + \frac{5}{16} y + \frac{3}{16} z + \frac{4}{16} u

(x + 7 - 2) (x + 7 + 2)

What is expand (-e^tsin(t)+cos(t)e^t)^2 ?
The solution to expand (-e^tsin(t)+cos(t)e^t)^2 is e^{2t}(-sin(2t)+1)