Answer to - proving $\cos 207^o$ is irrational.

Question

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.

Answer 1

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:

$$\cos(54°)=\sin(36°)=\sqrt{\frac{5-\sqrt{5}}{8}} \notin \Bbb{Q}$$

:)