Simplifying Algebraic Expression: A(Ls)

  • Thread starter Thread starter nesan
  • Start date Start date
  • Tags Tags
    Algebra
AI Thread Summary
The discussion focuses on simplifying the algebraic expression A(Ls) = (L - 2*Ls)*Ls*sin(beta) + Ls*sin(beta)*Ls*cos(beta). The simplification process involves eliminating brackets and reorganizing terms. The expression is transformed into A(Ls) = sin(beta)*[cos(beta) - 2]*Ls^2 + L*sin(beta)*Ls. Key steps include recognizing that Lb = L - 2*Ls, h = Ls*sin(beta), and w = Ls*cos(beta). The simplification emphasizes the importance of rearranging and combining like terms without using additional formulas.
nesan
Messages
74
Reaction score
0

Homework Statement



Sorry, not really a homework question, just practicing for a test online.

Can someone explain how this person simplified this line.

Homework Equations



A(Ls) = (L - 2*Ls)*Ls*sin(beta) + Ls*sin(beta)*Ls*cos(beta)

becomes

A(Ls) = sin(beta)*[cos(beta) - 2]*Ls^2 + L*sin(beta)*Ls

**INFO**

Lb = L - 2*Ls
h = Ls * sin(beta)
w = Ls * cos(beta)
A = Lb * h + 2 * 1/2 * h * w

Thank you, would be great help. :)
 
Physics news on Phys.org
He doesn't use any formulas at all. The only thing he is doing is changing the place of the brackets.

Hint: Start with eliminating the brackets in the term (L - 2*Ls)*Ls*sin(beta).
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top