<s>[INST] Answer to - proving $\cos 207^o$ is irrational. At first, $\cos(207^o) = \cos(180^o+27^o) = - \cos(27^o)$
We have, $(\sin 27^o + \cos 27^o)^2 = \sin^2 27^o + \cos^2 27^o + 2 \sin 27^o \cos 27^o$
$$ = 1 + \sin 2.27^o  = 1 + \sin 54^o = 1 + \sin (90^o - 36^o) = 1 + \cos 36^o$$
$$(\sin 27^o + \cos 27^o)^2 = 1 + \frac{\sqrt{5}+1}{4} = \frac{1}{4}(5+\sqrt 5)$$
$$\sin 27^o +  \cos 27^o = \frac{1}{2}{\sqrt{5+\sqrt 5}} ...(a)$$
Similarly, we have, 
$$(\sin 27^o- \cos 27^o)^2 = 1 - \cos 36° = 1 -\frac{\sqrt 5 +1}{4} = \frac{1}{4}(3-\sqrt 5)$$
$$ \sin 27^o - \cos 27^o = \pm\frac{1}{2}\sqrt{3-\sqrt 5}$$
Until $45^o$ in the first quadrant, $\sin x < \cos x$.
Thus,
$$ \sin 27^o - \cos 27^o = -\frac{1}{2}\sqrt{3-\sqrt 5}...(b)$$
On subtracting (a) from (b)
$$\cos 207^o = -\cos 27^o = - \frac{1}{4}(\sqrt{5+\sqrt5} + \sqrt{3 - \sqrt 5})$$
As $\sqrt{5+\sqrt5}$ and $\sqrt{3 - \sqrt 5}$ are irrational, $\cos 207^o$ is irrational.
 [/INST] As you noticed we have to prove that:
$$\cos(27°)\notin \Bbb{Q}$$
First of all:
$$ \cos(54°)=2\cos^2(27°)-1$$
So: $$\cos(54°)\notin \Bbb{Q}\Rightarrow \cos(27°)\notin \Bbb{Q}$$
Notice that:
$$\cos(54°)=\sin(36°)=\sqrt{\frac{5-\sqrt{5}}{8}} \notin \Bbb{Q}$$
:)
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