text: # Trig And calculus
  
  1. Oct 25, 2004
  
  ### alamin
  
  Need help in integration and trig
  How do you integrate
  1. 1/(1-x^5)
  2. 1/(1+x^4)
  
  and the trig question.
  Show that
  
  (a^2 - b^2)/c^2 = sin(A-B)/sin(A+B)
  
  2. Oct 25, 2004
  
  ### HallsofIvy
  
  Every polynomial, such as 1- x5 and 1+ x4 can be factored, using real numbers, into a product of linear or quadratic terms.
  To factor 1- x5, find all complex roots to x5= 1. One is, of course 1 the others are complex conjugates which can be paired to give two quadratic factors. Then use "partial fractions".
  Same for 1/(1+x4).
  
  In the trig question, are we to assum that a, b, and c are lengths of sides opposite angles A, B, C? In a right triangle or general triangle?
  
  3. Oct 26, 2004
  
  ### alamin
  
  general triangle!
  a,b,c are length's
  A,B,C are opposite angles
  
  Sorry but i made a mistake in the first integration question
  its suppose to be 1/sqrt(1-x^5)
  
  When i used this integration in mathematica 5 : i got something like hypergeometric2f1....
  
  Can u help me out!
  
  4. Oct 26, 2004
  
  ### HallsofIvy
  
  The 5 "fifth roots of unity" lie on a circle, in the complex plane of radius 1, equally spaced around the circle. The angle between them is 360/5= 72 degrees so they are;
  1, cos(72)+ i sin(72), cos(144)+ i sin(144), cos(216)+ i sin(216), cos(288)+ i sin(288).
  
  Since cos(72)= cos(288), sin(72)= -sin(288), cos(144)= cos(216), and sin(144)= sin(216), these are in pairs of complex conjugates (as they have to be in order to satisfy and equation with real coefficients.
  
  The solutions to x5= 1 are: 1, cos(72)+ i sin(72), cos(72)- i sin(72), cos(144)+ i sin(144), cos(144)- i sin(144) and so
  1- x= -(x-1)(x- cos(72)+ i sin(72))(x- 72- i sin(72))(x- cos(144)+ isin(144))(x- cos(144)- i sin(144))= -(x-1)((x-cos(72))2+ sin2(72))((x-cos(144)2+ sin2(144))
  
  = -(x-1)(x2- 2cos(72)+ 1)(x2-2cos(144)+ 1).
  
  Once you have that factorization you can expand 1/(1- x5) in partial fractions.