What is expand (e^ucos(u)-e^usin(u))^2 ?

The solution to expand (e^ucos(u)-e^usin(u))^2 is e^{2u}(-sin(2u)+1)

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expand (e^ucos(u)-e^usin(u))^2

expand

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

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

Solution steps

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

Apply Perfect Square Formula:

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

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

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

Simplify

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

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

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

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

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

= e^{2 u} (- sin (2 u) + 1)

- \frac{ln ( \frac{a ( x - b )}{s} + 1 )}{a + 1} - ln (s)

A (1, - 3) B (- 2, 1)

\frac{5 ( 13 + 6049 )}{49}

(5 - 3 i)^{6}

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

What is expand (e^ucos(u)-e^usin(u))^2 ?
The solution to expand (e^ucos(u)-e^usin(u))^2 is e^{2u}(-sin(2u)+1)