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)

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

\frac{5}{( x ^{2} + 4 ) ( x ^{2} - 1 )}

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

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

(3 z - \frac{2}{z})^{7}

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)