Given that $\log \sin x + \log \cos x = -1$ and that $\log (\sin x + \cos x) = 1/2(\log (n-1)$, find $n$.

Question

Given that $\log(\sin x) + \log(\cos x) = -1$ and that $\log (\sin x + \cos x) = 1/2(\log (n) - 1)$, find $n$.

I have no clue where to even start with this.

Answer 1

\begin{align} \log(\sin x) + \log(\cos x) &= -1\tag{1}\\ \log (\sin x + \cos x) &= \frac12\log (n-1)\tag{2} \end{align}

From $(1)$, $$\log(\sin x \cos x) = -1$$ then $$\sin x \cos x = e^{-1}$$

So you have

\begin{align} \log (\sin x + \cos x) &= \frac12\log (n-1)\\ 2\log (\sin x + \cos x) &= \log (n-1)\\ \log (\sin x + \cos x)^2 &= \log (n-1)\\ (\sin x + \cos x)^2&=n-1\\ \sin^2x+\cos^2x+2\sin x\cos x &=n-1\\ 1+2e^{-1}&=n-1\\ \end{align}