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

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

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

expand

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

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

Solution steps

(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 = e^{t} cos (t)

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

Simplify

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

\frac{11}{( 2 s + 5 ) ^{4}}

(\frac{1}{2} + \frac{2 a ^{2}}{b})^{2}

\frac{1}{( 2 + j w ) ^{2} ( 4 + j w )}

(d + e)^{3}

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