Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**miran****Guest**

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.

**ryos****Member**- Registered: 2005-08-04
- Posts: 394

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...

El que pega primero pega dos veces.

Offline

**Atled****Member**- Registered: 2005-08-22
- Posts: 9

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.

*Last edited by Atled (2005-09-14 08:50:32)*

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

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?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**miran****Guest**

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.

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

I'm very interested in knowing what c stands for?

**igloo** **myrtilles** **fourmis**

Offline

**miran****Member**- Registered: 2005-09-14
- Posts: 1

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.

Offline

Pages: **1**