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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

expand

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

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

Solution steps

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

Apply Perfect Square Formula:

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

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

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

Simplify

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

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

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

- (x + x \cdot y) d y

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

\frac{( 2 s + 7 + 6 e ^{- 4 s} )}{( s + 2 ) ( s + 6 )}

(cos (x) - sin (x)) \cdot (sin (x) - cos (x))

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