What is expand (-sin(t)e^t+cos(t)e^t)^2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand (-sin(t)e^t+cos(t)e^t)^2

expand

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

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

Solution steps

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

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

Apply Perfect Square Formula:

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

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

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

Simplify

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

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

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

(3 x + 5) (x - 25)

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

(e - f k) (e^{2} - e f k + f k^{2})

lo g_{10} (x) (x^{3} - 2)

[e^{- \frac{t}{2}} \cdot u \cdot (t - 2)]

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