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ChadG.1-Visitor
1-Visitor
June 16, 2011
Solved

Symbolic trignometric functions

  • June 16, 2011
  • 1 reply
  • 4760 views

How to tell the symbolic processor that n is an integer so while evaluating

cos((n+.5)* pi)==>0

and sin((n+.5)*pi)==>1

is there a way to force the symbolic process to do that once it sees these expressions it will just replace them with 0 or 1

The reason I am not using n as integer range is: this is part of a long calculation/derivation

Best answer by RichardJ

Add a "simplify"

1 reply

A
AlanStevens19-Tanzanite
19-Tanzanite
June 16, 2011

Chad G. wrote:

How to tell the symbolic processor that n is an integer so while evaluating

cos((n+.5)* pi)==>0

and sin((n+.5)*pi)==>1

is there a way to force the symbolic process to do that once it sees these expressions it will just replace them with 0 or 1

The reason I am not using n as integer range is: this is part of a long calculation/derivation

You might want to think more carefully about the values you've suggested for the cosines and sines! However, try using both the 'assume integer=n' and 'simplify' symbolic qualifiers.

Alan

C
ChadG.1-VisitorAuthor
1-Visitor
June 16, 2011

Thanks Alan I did not know about that "how to use the qualifiers assume n=integer?"

You think there is something wrong mathematically!

A
AlanStevens19-Tanzanite
19-Tanzanite
June 16, 2011

Chad G. wrote:

You think there is something wrong mathematically!

Well, cos((n+5)*pi) is -(-1)^n when n is an integer and sin((n+5)*pi) is 0.

Alan

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