Thread Subject: cos(pi/2) or sin(-pi) problem

hi guys..
i wanna ask you somthing..
i m using MATLAB R2008a and i have a problem with it..


matlab returns " cos(pi/2) = 6.1232e-017" and also " sin(-pi) = -1.2246e-016 ".. and this make me crazy.. both must be zero..

do you know how to fix this without using a "for" or "if" loop.. i mean to fix from the MATLAB program..
thanx..

muhaha <muhalkan@yahoo.com> wrote in message <15244285.1232199505643.JavaMail.jakarta@nitrogen.mathforum.org>...
> hi guys..
> i wanna ask you somthing..
> i m using MATLAB R2008a and i have a problem with it..
>
>
> matlab returns " cos(pi/2) = 6.1232e-017" and also " sin(-pi) = -1.2246e-016 ".. and this make me crazy.. both must be zero..
>

I'm not sure why it makes you crazy. You must know that machine math is often not exact. This is especially true when it involves numbers like pi, which have to be approximated using a finite number of digits and when it involves functions like cos() and sin() which can only be approximated by arithmetic operations.

Nevertheless, you might consider the following if you need to exact zeros.

>> cosd(90)

ans =

     0

>> sind(-180)

ans =

     0

Matt wrote:
> muhaha <muhalkan@yahoo.com> wrote in message <15244285.1232199505643.JavaMail.jakarta@nitrogen.mathforum.org>...
>> matlab returns " cos(pi/2) = 6.1232e-017" and also " sin(-pi) = -1.2246e-016 ".. and this make me crazy.. both must be zero..
>
> I'm not sure why it makes you crazy. You must know that machine math is often not exact. This is especially true when it involves numbers like pi, which have to be approximated using a finite number of digits and when it involves functions like cos() and sin() which can only be approximated by arithmetic operations.

That's right, but just to be clear: There is absolutely NOTHING wrong here. The irrational constant that mathematicians denote by the greek letter pi cannot be represented exactly in floating point, and the sin/cos above are the correct values for the floating point numbers that one gets when one types pi/2 and -pi at the MATLAB command line.

> Nevertheless, you might consider the following if you need to exact zeros.
>
>>> cosd(90)
>
> ans =
>
> 0
>
>>> sind(-180)
>
> ans =
>
> 0

Correct. 90 and -180 have an exact floating point representation.

Peter Perkins <Peter.PerkinsRemoveThis@mathworks.com> wrote in message <gkvt2k$4ab$1@fred.mathworks.com>...
>
> That's right, but just to be clear: There is absolutely NOTHING wrong here. The irrational constant that mathematicians denote by the greek letter pi cannot be represented exactly in floating point, and the sin/cos above are the correct values for the floating point numbers that one gets when one types pi/2 and -pi at the MATLAB command line.
> ......

  Strictly speaking that isn't quite true, Peter. It is impossible for 'cos' and 'sin' to return the exactly correct values for angles which are only rational approximations to pi/2 and pi, respectively, because the exact answers would themselves be irrational. For radian measure, except for x = 0, I believe it is true that there is no case where x and cosine(x), or x and sine(x), are simultaneously rational numbers.

Roger Stafford

If you must work with radians you might consider using a tolerance.

tol = eps; % for example.

num = [sin(-pi) cos(pi/2)];
num(abs(num)<eps)=0;




xxmm'bhga dZ3h\Z^^RxxnxF[m^ixhbi9^nagx`Z[xf\Zhohfa>ZehrZeZ

Roger Stafford wrote:

> Strictly speaking that isn't quite true, Peter. It is impossible for 'cos' and 'sin' to return the exactly correct values for angles which are only rational approximations to pi/2 and pi, respectively, because the exact answers would themselves be irrational.

Sure. I mean "correct" in the sense that those values are the correct floating point values for the sin and cos of the floating point approximations to pi/2 and -pi.

instead you can try this sin(sym(-pi)) or cos(syms(pi/2))...
the reason for cos(pi/2) = 6.1232 × 10^−17 is matlab approximates pi to 15 places after decimal point and sym(pi/2) is symbolic representation of pi/2 which gives you the exact answer

instead you can try this sin(sym(-pi)) or cos(syms(pi/2))...
the reason for cos(pi/2) = 6.1232 × 10^−17 is matlab approximates pi to 15 places after decimal point and sym(pi/2) is symbolic representation of pi/2 which gives you the exact answer

Dear Matt,

> xxmm'bhga dZ3h\Z^^RxxnxF[m^ixhbi9^nagx`Z[xf\Zhohfa>ZehrZeZ

An astonished armadillo has entered your keybord. Offer it some cashew nuts, then it will not eat up the Z.

Jan

"Jan Simon" <matlab.THIS_YEAR@nMINUSsimon.de> wrote in message <i9snk8$dmh$1@fred.mathworks.com>...
> Dear Matt,
>
> > xxmm'bhga dZ3h\Z^^RxxnxF[m^ixhbi9^nagx`Z[xf\Zhohfa>ZehrZeZ
>
> An astonished armadillo has entered your keybord. Offer it some cashew nuts, then it will not eat up the Z.
>
> Jan

Actually somebody solved this crypto-puzzle long ago. Look how old this thread is! I don't know why it was revived...

Dear Matt,

and my armadillo theories are not really new also.

Jan

I also noted. I do not have the symbolic toolbox. Is it the only possible solution?