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Maximum of...

by AlbertEinstein
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AlbertEinstein
#1
Jun23-06, 11:58 PM
P: 113
The question is find the maximum value of the following function

f(x) = 3cos(4*pi*x-1.3) + 5cos(2*pi*x+0.5).
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HallsofIvy
#2
Jun24-06, 09:22 AM
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PF Gold
P: 39,682
"The question"? Where, in your homework?

Looks pretty straight forward to me: differentiate and set the derivative equal to 0. Use the sin(a+b) formula to isolate [itex]sin(3\pi x)[/itex] and [itex]cos(3\pi x)[/itex].
AlbertEinstein
#3
Jun25-06, 02:06 AM
P: 113
Hey HallsofIvy,

Your suggestion was quite correct. However I dare say differentiating and equating to zero does not always help.
Here f (x) = 3cos (4*pi*x-1.3) + 5cos (2*pi*x+0.5)
Or, f (x) = - [12*pi*sin (4*pi*x-1.3) + 10*pi*sin (2*pi*x+0.5)]
Equating this to zero,
12*pi*sin (4*pi*x-1.3) + 10*pi*sin (2*pi*x+0.5)] = 0
Now if I use the sin (a+b) formula the equation gets rather complicated.

I shall be thankful to you if you could please show the full solution.
However the answer is 5.7811.

HallsofIvy
#4
Jun25-06, 12:16 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682
Maximum of...

Yes, it gets complicated. Anything wrong with that? You'll also, by the way, need to use the double angle formulas to reduce [itex]4\pi x[/itex] to [itex]2\pi x[/itex]. After you done all that, you will have an equation of the form [itex]A sin(2\pi x)+ B cos(2\pi x)= 0[/itex] with rather complicated numbers for A and B. But they are only numbers! Write [itex]tan(2\pi x)= -B/A[/itex] and solve.


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