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

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

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

cot (x) (cos (x) - sec (x))

R (- 4, 3, - 7) y S (0, 3, 0)

lo g_{10} (x^{5} \cdot y^{- 8})

d x (x^{2} - 3 x + 2) (2 x^{3} + 1)

(3 + 2 i)^{4}

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