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For $$0 < \theta < \dfrac{\pi}{4}$$, let
$$a = ((\sin \theta)^{\sin \theta}) (\log_{2} \cos \theta)$$, $$b = ((\cos \theta)^{\sin \theta}) (\log_{2} \sin \theta)$$
$$c = ((\sin \theta)^{\cos \theta}) (\log_{2} \cos \theta)$$ and $$d = ((\sin \theta)^{\sin \theta}) (\log_{2} \sin \theta)$$
Then, the median value in the sequence a, b, c, d is
We know when $$0 < \theta < \frac{\pi}{4}$$, $$\cos\theta\ >\sin\theta\ $$
Also, all the terms $$a,b,c,d$$ are multiplication of an exponential term (which is always positive) and a logarithmic term either $$\log_2\cos\theta\ $$ or $$\log_2\sin\theta\ $$
Now, both $$\sin\theta\ $$and $$\cos\theta\ $$ are fractional values in $$0 < \theta < \dfrac{\pi}{4}$$ range
So, both $$\log_2\cos\theta\ $$ and $$\log_2\sin\theta\ $$ will definitely be negative
So, $$a,b,c,d$$ all are negative terms
Now, let's compare $$a,b,c,d$$
i.) $$a$$ and $$d$$:
$$\dfrac{a}{d}=\dfrac{((\sin\theta)^{\sin\theta})(\log_2\cos\theta)}{((\sin\theta)^{\sin\theta})(\log_2\sin\theta)}$$
or, $$\dfrac{a}{d}=\dfrac{(\log_2\cos\theta)}{(\log_2\sin\theta)}$$
Since, $$\cos\theta\ >\sin\theta\ $$
or, $$\log_2\cos\theta\ >\log_2\sin\theta\ $$ (as $$\log_ax$$ is an increasing function for $$a>1$$)
or, $$\dfrac{\log_2\cos\theta}{\log_2\sin\theta}<1$$ (as both $$\log_2\cos\theta\ $$ and $$\log_2\sin\theta\ $$ are negative)
or, $$\dfrac{a}{d}<1$$
or, $$a>d$$ (as $$a$$ and $$d$$ both are negative) ------>(1)
ii.) $$b$$ and $$d$$:
$$\dfrac{b}{d}=\dfrac{((\cos\theta)^{\sin\theta})(\log_2\sin\theta)}{((\sin\theta)^{\sin\theta})(\log_2\sin\theta)}$$
or, $$\dfrac{b}{d}=\dfrac{((\cos\theta)^{\sin\theta})}{((\sin\theta)^{\sin\theta})}=\left(\cot\ \theta\ \right)^{\sin\ \theta\ }$$
Now for $$0 < \theta < \frac{\pi}{4}$$, $$\cot\theta\ $$ greater than 1 always
So, $$\dfrac{b}{d}>1$$
or, $$d>b$$ (as $$b$$ and $$d$$ both are negative) ------->(2)
iii.) $$a$$ and $$c$$:
$$\dfrac{a}{c}=\dfrac{((\sin\theta)^{\sin\theta})(\log_2\cos\theta)}{((\sin\theta)^{\cos\theta})(\log_2\cos\theta)}$$
or, $$\dfrac{a}{c}=\left(\sin\theta\ \right)^{\sin\ \theta\ -\cos\ \theta\ }$$
Now, $$\sin\theta\ -\cos\theta\ $$ is a negative term and $$\sin\theta$$ has a fractional value
So, $$\left(\sin\theta\right)^{\sin\theta\ -\cos\theta\ }$$ will definitely be greater than 1 (as fractional term having negative power is always greater than 1)
So, $$\dfrac{a}{c}>1$$
or, $$c>a$$ (as both a and c are negative) -------->(3)
So, from (1),(2) and (3),
$$c>a>d>b$$
So $$a,b,c,d$$ are now arranged in descending order.
Also, there are 4 terms in total.
So, median = average of $$3^{rd}$$ term and $$4^{th}$$ term = $$\dfrac{a+d}{2}$$
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