What is expand (-e^{-t}cos(t)-e^{-t}sin(t))^2 ?

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

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

lo g_{10} (x^{3} y^{- 4})

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

\frac{1}{( n + 1 ) ( n + 2 )}

(x + 3) (x - 3)

\frac{( x - 8 ) ( x + 5 )}{2}

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