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

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

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expand (e^xsin(x)-e^xcos(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 + x + 4) (2 - x - x + 4)

2 x ln (3 x) + (9 x^{2} + 2)^{3}

\frac{1}{( n - 1 ) n}

\frac{- x ^{2} + 3}{( x ^{2} + 3 ) ^{2}}

x^{x - 1} (ln (x) + \frac{x - 1}{x})

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