Height of Peak in M-Slit Experiments

[baubletop]

Homework Statement

Let q = kd sin(θ), so the sin(θ)-dependent factors in the M-slit and the 2-slit formulas for registered diffraction intensity can be compactly written:
1/M² = (sin²(Mq/2))/(sin²(q/2)) and cos²(q/2),
where the 1/M² prefactor normalizes them in common to have unit maximum values, to assist comparison. The first has its largest value when the denominator vanishes at q/2 = nπ and the second is largest where the cosine is maximum, at q/2 = nπ, so their maxima coincide.

A. What is the height of the peak following a maximum peak, for M=8?
B. Find the distance (in q units) beetween those peaks.

Homework Equations

See Above

The Attempt at a Solution

I'm having a very hard time trying to visualize this at all... My professor drew the graph of the functions (intensity + cosine) together, but this question is focusing more on the intensity equation.
Would I find the height of the smaller peak by taking the derivative of that function and setting it equal to 0? That's the best idea I have but I'm not sure if I'm totally off-base here. I feel like once I have part A I can do part B.

 

[lightgrav]

I think it's plug-and-chug: you are given M;
you already know that the max is at theta = 0 , (what is its n? what is its q?);
the "following peak" occurs for the next integer n+1 (what is its q?).

 

[baubletop]

When theta = 0, n = 1, so q = 2pi. The next "big" peak would be at the next integer value of n, but the question is asking about the smaller peaks between those large peaks (which don't occur at all in the cosine function). I can see them in the graph but I'm not sure how I would find their values from the given function without differentiating.

 

[lightgrav]

θ=0 means sinθ=0, so q=0, so n=0.
the next maximum happens at n=1, so what is its q?
what is its intensity?

 

[baubletop]

Ah right, my mistake. So at n = 1, q = 2pi (so q/2 = pi)? But then that gives an indeterminate form for the bigger equation. For cos it gives 1. This doesn't make much sense to me though.