Question

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 $n.$
(correct answer + 3, wrong answer 0)

Answer 1

${\log }_{10}\sin x+{\log }_{10}\cos x={\log }_{10}\left(\sin x\cos x\right)=-1$
Therefore, $\sin x\cos x=\frac{1}{10}\cdots \left(1\right)$
Now, manipulate the second equation.
${\log }_{10}(\sin x+\cos x)=\frac{1}{2}({\log }_{10}n-{\log }_{10}10)$
$\Rightarrow {\log }_{10}(\sin x+\cos x)={\log }_{10}\sqrt{\frac{n}{10}}$
$\Rightarrow \sin x+\cos x=\sqrt{\frac{n}{10}}$

$\Rightarrow (\sin x+\cos x{)}^2=\frac{n}{10}$

$\Rightarrow {\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x=\frac{n}{10}$
Substituting the value for $\sin x\cos x$ from $\left(1\right),$
$1+2\left(\frac{1}{10}\right)=\frac{n}{10}$
$\Rightarrow n=12$