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- miran
- 2005-09-15 19:22:02
I don´t know how to explane it. When i put in Mathematica a and b 45deg (a=b=45deg)
the function is plot (on x axis there´s c and on y axis amplitude) periodical from 0 to 1
in amplitude. Any other combination of a and b for example a=b=20deg will also achive
maximum at 1 but minimum would be in 0.6. The same is if the anyone of a or b is negative.
All this results are good. But if the a and b are not equal the ploted function would not have maximum at 1 but less then 1 and this solution is no good.
The question is how can i calculate from the 1/4(2+Cos[2(a-b)]+Cos[2(a+b)]-2Cos[c]Sin[2a]Sin[2b] a and b teoreticaly because upper results are only practical.
The real solution must be quide easy but i don´t see it.
- John E. Franklin
- 2005-09-15 17:02:17
I'm very interested in knowing what c stands for?
- miran
- 2005-09-15 16:36:08
At first i woul like to thank you for your help.
I apologize the corect function is:
1/4(2+Cos[2(a-b)]+Cos[2(a+b)]-2Cos[c]Sin[2a]Sin[2b])
i have simulated this function in program Mathematica and the right solutions seems to be a=b and a=-b or -a=b. Thats experimentaly, but i don´t know how to prove it matematicaly.?
It`s conected with my diploma research work in optical sensors.
- MathsIsFun
- 2005-09-15 07:26:18
miran wrote:
1/4(2+Cos[2(a-b)]+Cos[2(a+b)]-2Cos[c]Sin[a]Sin)
I need to know what the a and b should be that this function c would be maximum.
If you want c to be a maximum, and c is in the function, then we are missing something ... perhaps this function is equal to a constant?
Also if it is cos(c), then c could take on many values, as ryos mentioned already. For example (using radians) cos(1)=0.540, and cos(7.28)=0.540, etc, each time adding 2*pi.
You can also use Excel to plot the function to see how it behaves, and also try different a and b values. This may not get you an exact answer, but could help to point the way.
May I ask what this is related to?
- Atled
- 2005-09-15 06:47:13
1/4(2+Cos[2(a-b)]+Cos[2(a+b)]-2Cos[c]Sin[a]Sin) <- something seems wrong here
Sin needs a variable.
if a = 0 b= pi/4
you get 1/2
but you might be able to get something smaller than that.
You might try taking partial derivatives of Cos[2(a-b)]+Cos[2(a+b)] wrt a & b and look for extrema. You could also try to write an excel spread sheet which tries different combinations of a & b and use the goal seek function.
I am in a hurry but I will try to work on it later.
- ryos
- 2005-09-15 00:46:51
The cosine and sine functions are periodic, so they have infinitely many maximums. The period of both is pi radians (180 degrees), though cos begins pi/2 radians to the left of sin. So, at every increment of pi/2, either sin or cos has a maximum.
You can use this info to figure out the answer...
- miran
- 2005-09-14 20:41:19
hello
I have a function:
1/4(2+Cos[2(a-b)]+Cos[2(a+b)]-2Cos[c]Sin[a]Sin)
I need to know what the a and b should be that this function c would be maximum.