# Maximum of

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
Homework Helper
"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 $cos(3\pi x)$.

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.
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.