AMC 10 · 2017 · #7

Grade 11 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

periodic-functionfunction-compositiontrigonometric-ratios symmetry-argumentguess-and-check ↑ Prerequisites: periodic-functionfunction-composition

📏 Medium solution 💡 2 insights

Problem

The sine and cosine functions each repeat with least period two pi. Build the composite function f of x equal to cosine of sine x, and find its least period: the smallest positive number p with f of x plus p equal to f of x for every real x.

Pick an answer.

(A)

$\frac{\pi}{2}$

(B)

$\pi$

(C)

$2\pi$

(D)

$4\pi$

(E)

The function is not periodic.

How to solve

Strategy Organize Information in More Ways

The outer cosine is an even function, so it cannot tell the difference between an input and its negative. Rewriting cos(sin(x)) as cos(|sin(x)|) puts that fact up front: the composite only ever sees the size of sin(x), never its sign. Since |sin(x)| repeats twice as fast as sin(x) does, this rewrite immediately supplies a candidate period. Then the work splits in two: confirm the candidate really is a period for every x, and prove nothing smaller can be one by tracking where f reaches its maximum.

1STEP 1

Cosine ignores the sign

Cosine ignores the sign.

cos(-u)=cos(u) → cos(sin x)=cos(|sin x|)

2STEP 2

Shifting by pi does nothing

Shifting by pi flips only the inner sign, so the value is unchanged.

sin(x+π)=-sin x → f(x+π)=cos(-sin x)=cos(sin x)=f(x)

3STEP 3

Rule out the half-period pi/2

One test value kills the half period.

f(0)=cos 0=1, f (π/2)=cos 1≈ 0.5403 ≠ 1

4STEP 4

No shorter shift can work

Where the maximum occurs pins the least period at pi.

{x: f(x)=1}={x:sin x=0}={kπ} → p=kπ, p_min=π

Answer

Sample the function across one full turn: f(0) = 1, f(pi/4) = cos(0.7071) = 0.7602, f(pi/2) = cos 1 = 0.5403, f(3pi/4) = 0.7602, f(pi) = 1, f(5pi/4) = 0.7602, f(3pi/2) = 0.5403, f(7pi/4) = 0.7602, f(2pi) = 1. The list of values over [0, pi) repeats exactly over [pi, 2pi), which matches a period of pi. Meanwhile f(pi/2) differs from f(0), so pi/2 is not a period, and the values clearly do repeat, so the function is not aperiodic. Both facts agree with choice (B): every value repeats after pi and nothing shorter does.

💡Key takeaway

Cosine cannot tell an input from its negative, so cos(sin(x)) only ever sees |sin(x)|, which repeats every pi instead of every 2*pi.

Cosine ignores the sign
Shifting by pi does nothing
Rule out the half-period pi/2
No shorter shift can work

Cosine ignores the sign

Shifting by pi does nothing

Rule out the half-period pi/2

No shorter shift can work

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