View Full Version : Sinc(x) Pain
frenzie
May 19, 2004, 05:25 AM
Dear all,
Kindly help me with this final step.
How to simplify ((sin(x)/x)*cos(y)) - ((cos(x)-1)/x)*sin(y) to become
cos(y) only.
Thank you very much for easing my pain.
Sum
reinsuranc
Jun 24, 2004, 05:15 AM
If your equation is valid, then it is valid for any choice of x and y, other than values when the denominator is zero.
I picked x=pi, y=0. We know sin(pi)=0, cos(pi)=-1, sin(0)=0, cos(0)=1.
I get 0 on the left side of the equation , and 1 on the right side.
Your equation is not valid for this choice of x and y.
Dr_Calculus
Mar 25, 2005, 09:02 PM
You can't choose just any value for x and y since we have no knowledge of x and y being parts of a function, in which a specific x value would correspond to a specific y value.
reinsuranc
Mar 30, 2005, 08:52 AM
The original equation did not specify the range of values. In such a case, I think we can assume it is intended for all possible values of x and y, except those for which the denominator is zero.
It is possible that the equation was intended for a limited range of values, such as x and y each between PI/3 and PI/2, in which case my work provides no useful information, but it didn't say that.
If an equation is true for all values of x and y, except those for which the denominator is zero, then it must be true for any particular choice of x and y. So I made a particular choice of x and y. Since the equation did not hold true for that choice, I conclude that the equation is faulty.
Dr. Calculus, please explain why you think my logic is faulty. Thank you.
Dr_Calculus
Apr 9, 2005, 03:41 PM
I agree; I don't think that there's enough information to solve the problem unless there is some identity that I am unaware of... As for my criticism, I was just noting that we don't know whether y is a function of x, which may rule out choosing arbitrary values for both y and x. Anyway, the problem didn't explicitly say that (or at least not to our knowledge), so I guess it's not a significant comment.