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

Suma los logaritmos: la primera ecuación da

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

.

Add the logarithms: the first equation gives

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

Eleva al cuadrado

\sin x + \cos x

y usa

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

para evaluarlo.

Square

\sin x + \cos x

and use

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

to evaluate it

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.