Q11158: tan(pi/2) Does Not Cause an Error Because of Precision

Article: Q11158
Product(s): See article
Version(s): 3.00 4.00 5.00 5.10 6.00 6.00a | 5.10 6.00 6.00a
Operating System(s): MS-DOS | OS/2
Keyword(s): ENDUSER | s_quickc | mspl13_c
Last Modified: 15-JAN-1991

Question:

If you try to calculate tan(pi/2), which is in fact not defined, tan()
will return a value. This value is not large. Why isn't there an error
message for such an overflow?

Response:

This is a precision limitation. Because you are constrained to work in
single or double precision accuracy, it isn't possible to specify a
value close enough to the true value of pi/2 so that tan(X) overflows.

THE INFORMATION PROVIDED IN THE MICROSOFT KNOWLEDGE BASE IS PROVIDED "AS IS" WITHOUT WARRANTY OF ANY KIND. MICROSOFT DISCLAIMS ALL WARRANTIES, EITHER EXPRESS OR IMPLIED, INCLUDING THE WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE. IN NO EVENT SHALL MICROSOFT CORPORATION OR ITS SUPPLIERS BE LIABLE FOR ANY DAMAGES WHATSOEVER INCLUDING DIRECT, INDIRECT, INCIDENTAL, CONSEQUENTIAL, LOSS OF BUSINESS PROFITS OR SPECIAL DAMAGES, EVEN IF MICROSOFT CORPORATION OR ITS SUPPLIERS HAVE BEEN ADVISED OF THE POSSIBILITY OF SUCH DAMAGES. SOME STATES DO NOT ALLOW THE EXCLUSION OR LIMITATION OF LIABILITY FOR CONSEQUENTIAL OR INCIDENTAL DAMAGES SO THE FOREGOING LIMITATION MAY NOT APPLY.

Copyright Microsoft Corporation 1986-2002.