Logarithms and Trigonometry

Question

A multiple choice question asked the for the value of $\log_m(\sin x)$ given $\log_m(\cos x)=n$ where $\sin x$ and $\cos x$ are both positive and $m>1$. It is easily seen that the correct answer is $\cfrac{1}{2}\log_m(1-m^{2n})$. I was wondering if anyone could think of a possible extension to this short problem? I experimented with formulae for $\log_m(\tan x)$, $\log_m(\sin 2x)$ etc. but nothing interesting stood out.

Answer 1

$$\log_m(\cos x) = n \Rightarrow m^n = \cos x$$ $$m^{2n} + \sin^2 x = 1 \Rightarrow \sin x = \sqrt{1- m^{2n}} \Rightarrow \log_m(\sin x) = {1\over2}\log_m(1-m^{2n})$$

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$\sin 2x$

let $\log_m(2) = k$

$$\log_m(\cos x) + \log_m(\sin x) + \log_m(2) = n + {1\over2}\log_m(1-m^{2n}) + k$$ $$\log_m(\sin 2x) = n + {1\over2}\log_m(1-m^{2n}) + k$$

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$\tan x$

$${\log_m(\sin x) - \log_m(\cos x)} = {1\over2}\log_m(1-m^{2n}) - n$$ $$\log_m(\tan x) = {1\over2}\log_m(1-m^{2n}) - n$$

are these interesting or not ?