Max Value of f(x): 3cos(4πx-1.3) + 5cos(2πx+0.5)

[AlbertEinstein]

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).

 

[HallsofIvy]

"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 sin(3\pi x) and cos(3\pi x).

 

[AlbertEinstein]

Calculus isn't always helpful!

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]

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