AMC 10 · 2021 · #19

Grade 11 algebra
trigonometric-ratioscomplementary-anglesperiodic-functiondomain-restriction bound-inequality-then-enumerateconvert-to-algebra ↑ Prerequisites: trigonometric-ratios
📏 Long solution 💡 3 insights
Problem
Count the values in the closed interval from zero to pi that satisfy a certain trigonometric equation. The two sides are different trigonometric functions, and each one has another trigonometric function buried inside its angle.

Pick an answer.

(A)
0
(B)
1
(C)
2
(D)
3
(E)
4
How to solve
Strategy Organize Information in More Ways

You cannot compare a sine to a cosine directly, so Tool #15 (Organize Information in More Ways) does the load-bearing move: rewrite the cosine side as a sine using the cofunction identity, so both sides speak the same language. Normally "sin A = sin B" opens up infinitely many branches (A = B + 2π k and A = π - B + 2π k), which is where this problem eats most students. Tool #14 (Extreme Principle) closes that door first: check the extreme values of sin x and cos x on [0, π] and you find both inner angles are trapped inside [-π/2, π/2], the window where sine is one-to-one. Then the angles must simply be equal, Tool #13 (Convert to Algebra) reduces everything to sin x + cos x = 1, and Tool #2 (Make a Systematic List) reads off the finitely many solutions in the interval. Tool #3 (Eliminate Possibilities) is what kills the unwanted branches once the angles are trapped, and Tool #6 (Guess and Check) closes the job by substituting the two finalists back into the original equation.

1STEP 1

Trap both inner angles

Both inner angles sit in a narrow range.

π/2cos x ∈ [-π/2, π/2], π/2sin x ∈ [0, π/2]
2STEP 2

Rewrite the cosine as a sine

Put both sides in the same function.

cos(π/2sin x) = sin(π/2 - π/2sin x) ⟹ sin(π/2cos x) = sin(π/2 - π/2sin x)
3STEP 3

Cancel the sine safely

The range lets us cancel it safely.

A, C ∈ [-π/2, π/2] and sin A = sin C ⟹ π/2cos x = π/2 - π/2sin x
4STEP 4

Reduce to a clean equation

It reduces to sine plus cosine equals one.

cos x = 1 - sin x ⟹ sin x + cos x = 1
5STEP 5

Fold into one wave

Fold the sum into one sine wave.

√(2) sin(x + π/4) = 1 ⟹ sin(x + π/4) = √(2)/2
6STEP 6

List the solutions in range

List the solutions in range.

t = x + π/4 ∈ {π/4, 3π/4} ⟹ x = 0 or x = π/2
7STEP 7

Verify and count

Verifying gives 2.

x = 0: sinπ/2 = 1 = cos 0. x = π/2: sin 0 = 0 = cosπ/2. Count = 2 → (C)
Answer
2
Graph both sides on [0, π] and the count 2 is visible. The left side y = sin(π/2cos x) slides steadily down from 1 at x = 0, through 0 at x = π/2, to -1 at x = π. The right side y = cos(π/2sin x) starts at 1, dips to 0 at x = π/2, and climbs back to 1 at x = π. A strictly falling curve and a U-shaped curve that share their starting height and their midpoint height can only agree at those two places, and the endpoint x = π confirms it: the left side is -1 while the right side is 1, so no third meeting. The count also cannot be 0, since x = 0 obviously works, which rules out choice (A) immediately.
💡Key takeaway

When a sine is set equal to a cosine, first rewrite one as the other so both sides are the same function; then check whether both angles are trapped in [-π/2, π/2], because there the sine never repeats a value and the angles themselves must be equal. That collapses this monster into sin x + cos x = 1, which holds exactly at x = 0 and x = π/2 on [0, π] — so the answer is (C).

  • Trap both inner angles
  • Rewrite the cosine as a sine
  • Cancel the sine safely
  • Reduce to a clean equation
  • Fold the sum into one wave
  • List the solutions in range
  • Verify and count