## 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).
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 Recognitions: Gold Member Science Advisor Staff Emeritus "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. However the answer is 5.7811.

Recognitions:
Gold Member
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.