Find the value

For $z^{10} = 1\quad \Rightarrow e^{\frac{2k\pi}{10}i} = e^{36k^\circ i}$ , are the tenth roots of unity. From Argand's diagram, we have:

$\begin{aligned} \cos{72^\circ} +\cos{144^\circ} & = - \frac{1}{2} \\ \cos{72^\circ} + 2\cos^2{72^\circ} - 1 & = - \frac{1}{2} \\ 2\cos^2{72^\circ} + \cos{72^\circ} - \frac{1}{2} & = 0 \end{aligned}$

$\begin{aligned} \Rightarrow \cos{72^\circ} & = \frac{-1+ \sqrt{1+4}}{4} = \frac{\sqrt{5}-1}{4} \\ \Rightarrow \sin{72^\circ} & = \sqrt{ 1 - \left(\frac{\sqrt{5}-1}{4}\right)^2} = \sqrt{\frac{16 - 6+2\sqrt{5}}{16}} = \frac {\sqrt{10+2\sqrt{5}}}{4} \end{aligned}$

Now we have,

$\begin{aligned} \sin{27^\circ} & = \sin({72^\circ - 45^\circ}) = \frac{1}{\sqrt{2}} \sin{72^\circ} - \frac{1}{\sqrt{2}} \cos{72^\circ} \\ & = \frac {\sqrt{10+2\sqrt{5}}}{4\sqrt{2}} - \frac{\sqrt{5}-1}{4\sqrt{2}} = \frac {\sqrt{5 +\sqrt{5}}}{4} - \frac{\sqrt{(\sqrt{5}-1)^2}}{4\sqrt{2}} \\ & = \frac {\sqrt{5 + \sqrt{5}} }{4} - \frac{\sqrt{6-2\sqrt{5}}}{4\sqrt{2}} = \frac {\sqrt{5 +\sqrt{5}}}{4} - \frac{\sqrt{3-\sqrt{5}}}{4} \end{aligned}$

$\begin{aligned} \Rightarrow 4\sin{27^\circ} & = \sqrt{5 +\sqrt{5}} - \sqrt{3-\sqrt{5}} \\ \Rightarrow a + b & = 5 + 3 = \boxed{8} \end{aligned}$

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In response to the Challenge Master and Konstantinos Michailidis :

From $\begin{aligned} \sin{27^\circ} & = \sin({72^\circ - 45^\circ}) = \frac{1}{\sqrt{2}} \sin{72^\circ} - \frac{1}{\sqrt{2}} \cos{72^\circ} \end{aligned}$ , we have:

$\begin{aligned} 4\sin{27^\circ} & = 2\sqrt{2} ( \sin{72^\circ} - \cos{72^\circ}) \\ \Rightarrow \sqrt{a+\sqrt{a}} - \sqrt{b-\sqrt{a}} & = 2\sqrt{2} ( \sin{72^\circ} - \cos{72^\circ}) \\ \left(\sqrt{a+\sqrt{a}} - \sqrt{b-\sqrt{a}} \right)^2 & = 8 ( \sin^2{72^\circ} - 2 \sin{72^\circ} \cos{72^\circ}+ \cos^2{72^\circ}) \\ a + b - 2\sqrt{(a+\sqrt{a})(b-\sqrt{a})} & = 8(1-2 \sin{72^\circ} \cos{72^\circ}) \\ & = 8\left(1 - \frac{2\sqrt{10+2\sqrt{5}}(\sqrt{5}-1)}{16} \right) \\ & = 8 - \sqrt{2(5+\sqrt{5})}\sqrt{(\sqrt{5}-1)^2} \\ & = 8 - \sqrt{2(5+\sqrt{5})}\sqrt{6-2\sqrt{5}} \\ & = 8 - 2\sqrt{(5+\sqrt{5})(3-\sqrt{5})} \end{aligned}$

$\Rightarrow a = 5$ , $b = 3$ and $a+b = \boxed{8}$ .

$\begin{aligned}2\sin^{2}27^{\circ}&=1-\cos54^{\circ} \end{aligned}$

Note that $\cos54^{\circ}=\dfrac{\sqrt{10-2\sqrt{5}}}{4}$ .

Now, we have

$\begin{aligned}16\sin^{2}27^{\circ}&=8-2\sqrt{10-2\sqrt{5}}\\ 4\sin27^{\circ}&=\sqrt{8-2\sqrt{10-2\sqrt{5}}}=\sqrt{(x+y)-2\sqrt{xy}}=\sqrt{x}-\sqrt{y},~~x>y\end{aligned}$

By solving both of these equations, $x+y=8$ and $xy=10-2\sqrt{5}$ , we get $x=5+\sqrt{5}$ and $y=3-\sqrt{5}$ .

Finally,

$4\sin27^{\circ}=\sqrt{5+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{a+\sqrt{a}}-\sqrt{b-\sqrt{b}}$

Hence, $\large{\boxed{a+b=8}}$

Just use sin(a-b) formula where a=45,b=18. And put values of sin and cos of 18 degrees