What is expand (e^xcos(x)-sin(x)e^x)^2 ?

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

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expand (e^xcos(x)-sin(x)e^x)^2

expand

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

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

Solution steps

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

= (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 = sin (x) e^{x}

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

Simplify

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

2 (x^{2} + 2 x - 5)^{3}

\frac{( 8 x ^{2} + 13 )}{( x ^{2} + 1 ) ( x + 1 )}

1 + \frac{3.3 ( s + 6 ) ( s + 10.75 )}{( s + 2 ) ( s + 3 ) ( s + 5 )}

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