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

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

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

expand

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

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

Solution steps

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

Apply the distributive law:

a (b - c) = ab - a c

a = (2 e^{t} cos (t)), b = e^{t} cos (t), c = e^{t} sin (t)

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

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

Simplify

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

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

\frac{( 5 x - 22 ) ( x - 1 )}{x + 5}

\frac{\partial}{\partial x} ((y) (x^{3} y - x - 5 y - 10))

(x^{2} - 5)^{8} \cdot 2 x d x

\frac{( 4 - 3 i ) ^{6}}{6}

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

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