AMC 10 · 2007 · #17

Grade 11 algebra
pythagorean-identityangle-addition-formulatrigonometric-ratiosdouble-angle-formula formula-substitutionconvert-to-algebra ↑ Prerequisites: pythagorean-identitytrigonometric-ratios
📏 Long solution 💡 3 insights
Problem
Two angles have a known sum of sines and a known sum of cosines. Find the cosine of their difference.

Pick an answer.

(A)
$\sqrt{\frac{5}{3}} - 1$
(B)
$\frac 13$
(C)
$\frac 12$
(D)
$\frac 23$
(E)
1
How to solve
Strategy Organize Information in More Ways

Solving for a and b looks hopeless, and it is also unnecessary. Tool #13 (Convert to Algebra) opens with the subtraction formula, which shows the target is built from cos acos b and sin asin b — two products. The load-bearing move is Tool #15 (Organize Information in More Ways): stop reading the two givens as two separate facts and combine them, because squaring a sum manufactures exactly the cross-product hiding inside it, and adding the two squared equations lets the Pythagorean identity swallow everything that is not wanted. That settles what the value must be. It does not settle that anything has that value, because squaring only runs one way, so Tool #7 (Identify Subproblems) splits the job in two: the forced value, and the existence of a pair. Tool #4 (Introduce a Variable) sets up half-angle variables s=(a+b)/2 and d=(a-b)/2 in which the system untangles, and Tool #11 (Work Backwards) uses them to build an explicit pair and plug it back in. Tool #3 (Eliminate Possibilities) is the ten-second filter that kills choice (E) before any of this starts.

1STEP 1

See what the target is made of

The target expands into exactly the cross terms.

cos(a-b) = cos acos b + sin asin b
2STEP 2

Square each given sum

Squaring each given sum produces those cross terms plus squares.

(sin a+sin b)² = sin² a + sin² b + 2sin asin b = 5/3 (cos a+cos b)² = cos² a + cos² b + 2cos acos b = 1
3STEP 3

Add, and let the identity eat the squares

Adding lets the identity eat every square.

(sin² a+cos² a)₁+(sin² b+cos² b)₁+2(sin asin b+cos acos b) = 5/3+1 = 8/3
4STEP 4

Read off the forced value

Reading off gives 1/3.

2 + 2cos(a-b) = 8/3 ⟹ 2cos(a-b) = 2/3 ⟹ cos(a-b) = 1/3
5STEP 5

Untangle the system with half-angles

Half-angle forms untangle the original system.

sin a+sin b = 2sin scos d = √(5/3) and cos a+cos b = 2cos scos d = 1
6STEP 6

Build a pair that works

That builds an actual pair of angles.

cos² d = 2/3 ⟹ cos d = √(6)/3, sin d = √(3)/3; cos s = 1/(2cos d) = √(6)/4, sin s = 1/(2cos d)√(5/3) = √(10)/4
7STEP 7

Plug it back in and finish

Substituting back confirms 1/3, choice (B).

cos a+cos b = 12/12 = 1, sin a+sin b = 4√(15)/12 = √(5/3), cos(a-b) = 2cos² d-1 = 1/3 ⟹ (B)
Answer
1/3
Four independent checks agree. (1) Range: 1/3 lies in [-1,1], as any cosine must. (2) Choice (E) dies in one line without any of the work above: cos(a-b)=1 forces a≡ b, hence 2cos a=1 and 2sin a=√(5/3), giving cos² a+sin² a=1/4+5/12=2/3 ≠ 1. (3) Numerically, the constructed pair is a≈ 1.52722 and b≈ 0.29626 radians; then cos a+cos b≈ 0.04356+0.95644=1.00000, sin a+sin b≈ 0.99905+0.29194=1.29099=√(5/3), and cos(a-b)=cos(1.23096)≈ 0.33333. (4) Cross-check by the parallelogram law: subtracting instead of adding the squared equations style gives (cos a-cos b)²+(sin a-sin b)² = 2-2cos(a-b) = 4/3, and 8/3+4/3=4=2(1+1), exactly as the law demands. Two decoys are worth naming. Choice (D) 2/3 is precisely cos² (a-b)/2, the number you stop at if you forget the final double-angle step — it is the trap for the half-angle route, not a random distractor. Choice (A) √(5/3)-1 is the "combine the two given numbers" decoy. Finally, the pair is not merely existent but essentially unique: in the half-angle system cos d=±√(2/3), and each sign then forces s completely, with the two choices only swapping a and b. So {a,b} is determined modulo 2π, and even the individual cross-products are pinned: cos acos b=1/24 and sin asin b=7/24, whose sum is 8/24=1/3 — the answer recovered a fourth way, this time from the two products separately.
💡Key takeaway

Square both given sums and add them: the Pythagorean identity swallows all four squares into a plain 2, and what is left over is exactly 2cos(a-b).

  • See what the target is made of
  • Square each given sum
  • Add, and let the identity eat the squares
  • Read off the forced value
  • Untangle the system with half-angles
  • Build a pair that works
  • Plug it back in and finish