Question

Dado que $\log_{10} \sin x + \log_{10} \cos x = -1$ y que $\log_{10}(\sin x + \cos x)$ $= \frac{1}{2}(\log_{10} n - 1),$ halla $n.$

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 $n.$

Respuesta

Solución:

La primera ecuación dice $\log_{10}(\sin x \cos x) = -1,$ así que $\sin x \cos x = \frac{1}{10}.$ Entonces $\begin{aligned} (\sin x + \cos x)^2 &= \sin^2 x + \cos^2 x \\ &\quad {}+ 2 \sin x \cos x \\ &= 1 + \frac{2}{10} = \frac{12}{10}. \end{aligned}$

Tomando logaritmos, $2\log_{10}(\sin x + \cos x)$ $= \log_{10} \frac{12}{10}$ $= \log_{10} 12 - 1,$ así que $\log_{10}(\sin x + \cos x)$ $= \frac{1}{2}(\log_{10} 12 - 1)$ y $n = 12.$

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 $n = 12.$

2003 AIME I Problema 4

Solución:

El Problema 4 en otros años