Its sine function!

Starting from the identity $\cos \dfrac \pi 5 + \cos \dfrac {3\pi}5 = \dfrac 12$ ,

$\begin{aligned} \cos 36^\circ + \cos 108^\circ & = \frac 12 \\ \cos 36^\circ - \cos 72^\circ & = \frac 12 \\ 4\cos^2 36^\circ - 2\cos 36^\circ - 1 & = 0 \\ \implies \cos 36^\circ & = \frac {1+\sqrt 5}4 \\ 2 \cos^2 18^\circ - 1 & = \frac {1+\sqrt 5}4 \\ \implies \cos^2 18^\circ & = \frac {5+\sqrt 5}8 \\ \left(1-2\sin^2 9^\circ \right)^2 & = \frac {5+\sqrt 5}8 \\ 32 \sin^4 9^\circ - 32 \sin^2 9^\circ + 3 - \sqrt 5 & = 0 \\ \sin^2 9^\circ & = \frac {4 \blue - \sqrt{2(5+\sqrt 5)}}8 & \small \blue{\text{Note that }\sin 9^\circ \approx 0} \\ \implies \sin 9^\circ & = \sqrt{\frac {2-\sqrt{\sqrt 5 \cdot \frac {\sqrt 5 +1}2}}4} \\ & = \sqrt {\frac {2-\sqrt[2^2]5 \cdot \sqrt \blue \varphi}{2^2}} & \small \blue{\text{where }\varphi \text{ denotes the golden ratio.}} \end{aligned}$

Therefore $\log_a \left(\dfrac {80\sin(2ab^\circ)}b \right) = \log_2 \left( \dfrac {80 \sin 30^\circ}5 \right) = \log_2 8 = \boxed 3$ .

Claim: $\cos(36^\circ) = \dfrac{1+\sqrt5}4$ .

Proof : Using the complementary angle formula, $\cos(A) = - \cos(180^\circ - A)$ , we have $\cos(36^\circ) + \cos(144^\circ) = 0$

Apply the double-angle formula, $\cos(2A) = 2\cos^2(A) - 1$ .

With $\cos(144^\circ) = 2\cos^2(72^\circ) - 1$ and $\cos(72^\circ) = 2\cos^2(36^\circ) - 1$ . Also denote $y = \cos(36^ \circ)$ , $y + \Big[2 (2y^2 - 1)^2 - 1\Big] = 0 \quad\Leftrightarrow \quad 8y^4 - 8y^2 + y + 1 = 0 \quad\Leftrightarrow \quad (y + 1)(2y-1)(4y^2-2y - 1 ) = 0$

Since $1> \cos(36^\circ) > \cos(60^\circ) = \frac12$ , then $y = \cos(36^\circ) = \dfrac{1+\sqrt5}4$ . $\quad \blacksquare$

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Now we will be using the half-angle formula repeatedly, $\cos(A) = 2\cos^2(\tfrac A2) - 1 = 1 - 2\cos^2(\tfrac A2)$

Then, $\begin{array} { r c l } \cos(36^\circ) &=& \dfrac{1+\sqrt5}4 \\ 4 \cos^2(18^\circ) &=& \dfrac{\sqrt5 + 5}2 \\ 2\cos(18^\circ) &=& \sqrt[4]{5} \cdot \sqrt{\phi} \\ 2 - 4\sin^2(9^\circ) &=& \sqrt[4]{5} \cdot \sqrt{\phi} \\ \sin(9^\circ) &=&\large \sqrt{\dfrac{2 - \sqrt[4]{5} \cdot \sqrt{\phi}}{4} } \end{array}$ Hence, $(a,b) = (2,5) \Rightarrow \log_a \left(\dfrac{80 \sin(3ab\degree)}{b} \right) = \boxed3 .$

$2\sin(\frac{π}{20})\cos(\frac{π}{20}) = \sin(\frac{π}{10})$ Taking $\sin(\frac{π}{20}) = t , \cos(\frac{π}{20})= \sqrt{1-t^2 }$

$\blue\sin{(\frac{π}{10})}= \frac{\sqrt{5}-1}{4}$ $2t\sqrt{1-t^2} = \frac{\sqrt{5}-1}{4} \implies t^2 (1-t^2) = (\frac{\sqrt{5}-1}{8})^2$ $t^4 -t^2 +\frac{3-\sqrt{5}}{32}=0\implies t^2 = \frac{1±\sqrt{1-\frac{3-\sqrt{5}}{8}}}{2}$ $t= \sqrt{\frac{2-\sqrt[4]{5}\sqrt{\frac{\sqrt{5}+1}{2}}}{2}}$ $=\sqrt{\frac{2-\sqrt[4]{5}\sqrt{\phi}}{2}}$ Answer is $\color{#20A900} \boxed {\log_2 (\frac{80\sin(30°)}{5})=3}$