1975 AMC 12 Problem 30

Attempt Problem 30 of the 1975 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1975 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

30.

Let x=cos⁡36∘−cos⁡72∘.x=\cos36^\circ-\cos72^\circ. Then xx equals
13\frac13
12\frac12
3−63-\sqrt6
23−32\sqrt3-3
none of these
Answer: B
Concepts:trigonometric identityalgebraic manipulation
Small hint:
Let w=cos⁡36∘w=\cos36^\circ and y=cos⁡72∘y=\cos72^\circ
Big hint:
Use y=2w2−1y=2w^2-1 and w=1−2y2w=1-2y^2, then add the equations
Solution:
Set w=cos⁡36∘w=\cos36^\circ and y=cos⁡72∘.y=\cos72^\circ. The double-angle identities give y=2w2−1,w=1−2y2. y=2w^2-1,\qquad w=1-2y^2. Adding yields w+y=2(w2−y2)=2(w−y)(w+y). \begin{aligned} w+y&=2(w^2-y^2)\\ &=2(w-y)(w+y). \end{aligned} Since w+y≠0,w+y\ne0, division gives w−y=12.w-y=\frac{1}{2}. Thus x=12.x=\frac{1}{2}.
Therefore, the correct answer is B.
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