20050322, 21:48  #1 
Jan 2005
1101_{2} Posts 
please help me, cal expert
Find equivalent expressions( shortest is best) for the following do not involce any trig function or w or T. Thanks
EX: sin(n*pi) =0 for all n and cos(n*pi)=(1)^n for all n a) cos((2n+1)*pi/2) where n is a positive integer b) sin((2n+1)*w*T/2) where n is a positive integer c) cos(n*w*T/2) where n is a positive integer d) sin(2n+1)*pi/2)  cos(2n*pi/2) where n is a positive integer p/s: w is omega 
20050322, 23:05  #2 
Jan 2005
D_{16} Posts 
i got a) =0
any one know how to get b and c ? thanks 
20050323, 20:05  #3 
∂^{2}ω=0
Sep 2002
República de California
11,743 Posts 
I thought the whole point of homework was to figure it out for yourself...

20070924, 00:20  #4 
67·139 Posts 
I thought the whole point of this forum is to help people with math homework?

20070924, 16:54  #5 
"Nancy"
Aug 2002
Alexandria
2467_{10} Posts 
What are w and T? If those are just indeterminates, I think there's not much you can do to simplify b) and c).
Alex 
20070924, 17:19  #6  
"Bob Silverman"
Nov 2003
North of Boston
5^{2}×13×23 Posts 
Quote:
the first cardinal infinity. Now, since the domain of sin() and cos() is the real numbers and omega is not a real number, it makes the expression(s) gibberish. 

20070924, 17:20  #7 
"Phil"
Sep 2002
Tracktown, U.S.A.
3·373 Posts 

20070924, 17:23  #8  
"Bob Silverman"
Nov 2003
North of Boston
5^{2}·13·23 Posts 
Quote:
To get help, you first need to show what you have done already with each of the problems. Then we can guide you. 

20070924, 19:12  #9 
Aug 2002
Buenos Aires, Argentina
2×17×43 Posts 
w = 2*pi*f = 2*pi/T so it should be easy to continue from here. But I think this help is too late for him since he posted his questions in 2005.
Last fiddled with by alpertron on 20070924 at 19:14 
20070924, 19:42  #10 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 

20070924, 21:55  #11 
"Nancy"
Aug 2002
Alexandria
9A3_{16} Posts 
Or it was someone else wondering about Ernst's response. In fact, I think the Homework Help forum was formed after 2005, and this old thread moved there.
Alex 
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