2003 AIME I Problem 4

Attempt Problem 4 of the 2003 AIME I 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 2003 AIME I solutions, or check the answer key.

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4.

Given that

\log_{10} \sin x + \log_{10} \cos x = -1

and that

\log_{10}(\sin x + \cos x)

= \frac{1}{2}(\log_{10} n - 1),

find

n.

Answer

Answer: 12

Concepts:logarithm trigonometric identity

Difficulty rating: 1990

Small hint:

Add the logarithms: the first equation gives

\sin x \cos x = \frac{1}{10}

Big hint:

Square

\sin x + \cos x

and use

\sin^2 x + \cos^2 x = 1

to evaluate it

Solution:

The first equation says

\log_{10}(\sin x \cos x) = -1,

so

\sin x \cos x = \frac{1}{10}.

Then

\begin{aligned} (\sin x + \cos x)^2 &= \sin^2 x + \cos^2 x \\ &\quad {}+ 2 \sin x \cos x \\ &= 1 + \frac{2}{10} = \frac{12}{10}. \end{aligned}

Taking logarithms,

2\log_{10}(\sin x + \cos x)

= \log_{10} \frac{12}{10}

= \log_{10} 12 - 1,

so

\log_{10}(\sin x + \cos x)

= \frac{1}{2}(\log_{10} 12 - 1)

and

n = 12.