1974 AMC 12 Problem 22

Attempt Problem 22 of the 1974 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 1974 AMC 12 solutions, or check the answer key.

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

22.

The minimum value of sin⁡A2−3cos⁡A2\sin\frac A2-\sqrt3\cos\frac A2 is attained when AA is
−180∘-180^\circ
60∘60^\circ
120∘120^\circ
0∘0^\circ
none of these
Answer: E
Concepts:trigonometric identitytrigonometry
Small hint:
Combine the sine and cosine into one shifted sine with amplitude 22
Big hint:
Use sin⁡u−3cos⁡u=2sin⁡(u−60∘)\sin u-\sqrt3\cos u=2\sin(u-60^\circ)
Solution:
Let u=A2.u=\frac{A}{2}. Then sin⁡u−3cos⁡u\sin u-\sqrt3\cos u equals 2sin⁡(u−60∘).2\sin(u-60^\circ). Its minimum occurs when u−60∘=270∘+360∘n,u-60^\circ=270^\circ+360^\circ n, so A=660∘+720∘n A=660^\circ+720^\circ n for an integer n.n. None of the first four choices has this form.
Therefore, the correct answer is E.
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