Trigonometry Values Pi/10

$$\sin \left({\pi\over 10}\right) = \sin\left({{1\over 2}\cdot{\pi\over 5}}\right)= \sqrt{{\textstyle{1\over 2}}\left[{1-\cos\left({\pi\over 5}\right)}\right]}$$

$$= \sqrt{{\textstyle{1\over 2}}[1-{\textstyle{1\over 4}}(1+\sqrt{5}\,)]} = {\textstyle{1\over 4}}(\sqrt{5}-1).$$ (1)

So we have

$$\cos\left({\pi\over 10}\right) = \cos\left({{1\over 2}\cdot {\pi\over 5}}\right)=\sqrt{{1\over 2}\left[{1+\cos\left({\pi\over 5}\right)}\right]}$$

$$= \sqrt{{\textstyle{1\over 2}}[1+{\textstyle{1\over 4}}(1+\sqrt{5}\,)]}$$

$$= {\textstyle{1\over 4}}\sqrt{10+2\sqrt{5}},$$ (2)

and

$$\tan\left({ \pi\over 10}\right)= \sqrt{3-\sqrt{5}\over 5+\sqrt{5}} = {\textstyle{1\over 5}}\sqrt{25-10\sqrt{5}}.$$ (3)

Summarizing,

$$\sin \left({\pi\over 10}\right) = {\textstyle{1\over 4}}(\sqrt{5}-1)$$ (4)

$$\cos\left({\pi\over 10}\right) = {\textstyle{1\over 4}}\sqrt{10+2\sqrt{5}}$$ (5)

$$\tan\left({ \pi\over 10}\right) = {\textstyle{1\over 5}}\sqrt{25-10\sqrt{5}}$$ (6)

$$\sin\left({3\pi\over 10}\right) = {\textstyle{1\over 4}}(1+\sqrt{5}\,)$$ (7)

$$\cos\left({3\pi\over 10}\right) = {\textstyle{1\over 4}}\sqrt{10-2\sqrt{5}}$$ (8)

$$\tan\left({3\pi\over 10}\right) = {\textstyle{1\over 5}}\sqrt{25+10\sqrt{5}}.$$ (9)

An interesting near-identity is given by

$${1\over 4}\left[{\cos({\textstyle{1\over 10}})+\cosh({\texts... ...}\,)\cosh({\textstyle{1\over 20}}\sqrt{2}\,)}\right]\approx 1.$$ (10)

In fact, the left-hand side is approximately equal to $1+2.480\times 10^{-13}$.

See also Decagon, Decagram