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

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

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

expand

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

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

Solution steps

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

Apply Perfect Square Formula:

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

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

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

Simplify

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

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

4 (m x - m + 4)^{2}

e^{x + 0.6} - 2 \cdot (x + 0.7)^{2}

d (t (1 - t) (1 - 2 t)) + a (t (1 - t) (1 - 2 t))

\frac{( 12 v ^{3} + 38 v ^{2} - 16 v - 20 )}{( 6 v - 5 )}

\frac{- 3 x ^{2} - 4 x - 5}{x ( x + 1 ) ( x - 5 )}

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