AMC12 2007 A

AMC12 2007 A · Q17

AMC12 2007 A · Q17. It mainly tests Manipulating equations, Trigonometry (basic).

Suppose that $\sin a + \sin b = \sqrt{\frac{5}{3}}$ and $\cos a + \cos b = 1$. What is $\cos (a - b)$?

假设 $\sin a + \sin b = \sqrt{\frac{5}{3}}$ 且 $\cos a + \cos b = 1$。求 $\cos (a - b)$？

(A) $\sqrt{5}/3 - 1$ $\sqrt{5}/3 - 1$

(B) 1/3 1/3

(D) 2/3 2/3

(E) 1 1

Answer

Correct choice: (B)

正确答案：（B）

Solution

We can make use the of the trigonometric Pythagorean identities: square both equations and add them up: This is just the cosine difference identity, which simplifies to $\cos (a - b) = \frac{1}{3} \Longrightarrow \mathrm{(B)}$

将两个给定方程两边平方，得到 $\sin^2 a + 2 \sin a \sin b + \sin^2 b = 5/9$ 和 $\cos^2 a + 2 \cos a \cos b + \cos^2 b = 1$。将对应项相加，得到 $({\sin^2 a + \cos^2 a}) + ({\sin^2 b + \cos^2 b}) + 2(\sin a \sin b + \cos a \cos b) = 14/9$。因为 $\sin^2 a + \cos^2 a = \sin^2 b + \cos^2 b = 1$，所以 $\cos(a - b) = \sin a \sin b + \cos a \cos b = (14/9 - 2)/2 = 1/3$。

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