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From: harryos on 28 Oct 2009 05:07
hi
I was trying to calculate the value of
x(t)=cos(2*pi*10*t)+cos(2*pi*25*t)+cos(2*pi*50*t)+cos(2*pi*100*t) for
some values.I wrote this code

....
public static void main(String[] args) {
double m1=0.0;
double m2=0.0;
double m3=0.0;
double m4=0.0;
double y=0.0;
for(int t=0;t<3;t++){
debug("t="+t);
m1=Math.cos(Math.PI*2*10*t);
m2=Math.cos(Math.PI*2*25*t);
m3=Math.cos(Math.PI*2*50*t);
m4=Math.cos(Math.PI*2*100*t);
y=m1+m2+m3+m4;
System.out.println("m1="+m1);
System.out.println("m2="+m2);
System.out.println("m3="+m3);
System.out.println("m4="+m4);
System.out.println("y="+y);

}

}
....

however ,I got the following output
t=0
m1=1.0
m2=1.0
m3=1.0
m4=1.0
y=4.0

t=1
m1=1.0
m2=1.0
m3=1.0
m4=1.0
y=4.0

t=2
m1=1.0
m2=1.0
m3=1.0
m4=1.0
y=4.0
-----
I don't understand this..For example, the value of cos(2*pi*10*1)
should be cos 62.832 which is .4566 .Instead I am getting the value 1.
can somebody help me find out if I am doing something wrong?
thanks
harry
From: Arved Sandstrom on 28 Oct 2009 05:13
harryos wrote:
> hi
> I was trying to calculate the value of
> x(t)=cos(2*pi*10*t)+cos(2*pi*25*t)+cos(2*pi*50*t)+cos(2*pi*100*t) for
> some values.I wrote this code
[ SNIP ]

> I don't understand this..For example, the value of cos(2*pi*10*1)
> should be cos 62.832 which is .4566 .Instead I am getting the value 1.
> can somebody help me find out if I am doing something wrong?
> thanks
> harry

You'll get dozens of answers here, but I see none posted yet - the
argument is in radians, not degrees. So cos (20pi) = 1.

AHS
From: harryos on 28 Oct 2009 05:42
On Oct 28, 2:13 pm, Arved Sandstrom <dces...(a)hotmail.com> wrote:
the argument is in radians, not degrees. So cos (20pi) = 1.
> AHS

thanks for the reply..
if I use

m1=Math.cos(Math.PI*2*10*t*180/Math.PI)

will that do the conversion?

harry

From: Lars Enderin on 28 Oct 2009 05:47
harryos wrote:
> On Oct 28, 2:13 pm, Arved Sandstrom <dces...(a)hotmail.com> wrote:
> the argument is in radians, not degrees. So cos (20pi) = 1.
>> AHS
>
> thanks for the reply..
> if I use
>
> m1=Math.cos(Math.PI*2*10*t*180/Math.PI)
>
> will that do the conversion?
>
What's the point of multiplying, then dividing, with PI?
If t is in degrees, you need to multiply with PI/180 to get radians.
From: Andreas Leitgeb on 28 Oct 2009 08:06
harryos <oswald.harry(a)gmail.com> wrote:
> On Oct 28, 2:13 pm, Arved Sandstrom <dces...(a)hotmail.com> wrote:
> > the argument is in radians, not degrees. So cos (20pi) = 1.
> if I use m1=Math.cos(Math.PI*2*10*t*180/Math.PI)
> will that do the conversion?

If with 20pi you really thought of an angle in degrees,
(that would be slightly larger than the angles of an
equilateral triangle), then Math.PI*2*10*t*180/Math.PI
or just: 3600*t does it.

Usually it doesn't make any sense in real world to use
Pi-multiples (like 20pi), interprete them as degrees
and then convert them to radians to feed to trigonometric
functions, but if you're just playing around, then such a
sense is of course not a necessity.

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