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

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

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

expand

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

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

Solution steps

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

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

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

Expand

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

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

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

Factor out common term

e^{2 t}

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

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

(2 - x) e^{x}

4 (1 + e^{2})

2 y d x + (x - sin (y^{\frac{1}{2}})) d y

(a^{4}) 3

\frac{x + 8}{( x + 3 ) ^{2}}

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