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

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

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expand (-e^{-x}sin(x)-e^{-x}cos(x))^2

expand

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

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

Solution steps

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

Apply Perfect Square Formula:

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

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

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

Simplify

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

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

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

2^{x} ln (2) (x + 1)

\frac{x}{( x + 1 ) ( x - 4 )}

(x + 5)^{2} d x

\frac{s + 3}{( s + 1 ) ( s + 2 )}

(x^{2} + e^{x}) (2 x e^{x})

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