Trigonometry Values Pi/5

Use the identity

$$\sin(5\alpha)=5\sin\alpha-20\sin^3\alpha+16\sin^5\alpha.$$ (1)

Now, let $\alpha\equiv{\pi/5}$ and $x\equiv\sin\alpha$. Then

$$\sin\pi=0=5x-20x^3+16x^5$$ (2)

$$16x^4-20x^2+5=0.$$ (3)

Solving the Quadratic Equation for $x^2$ gives

$$\sin^2\left({\pi\over 5}\right) = x^2 = {{20\pm\sqrt{(-20)^2-4\cdot 16\cdot 5}}\over 2\cdot 16}$$

$$= {20\pm \sqrt{80}\over 32} = {\textstyle{1\over 8}}(5\pm\sqrt{5}\,).$$ (4)

Now, $\sin(\pi/5)$ must be less than

$$\sin\left({\pi\over 4}\right)= {\textstyle{1\over 2}}\sqrt{2},$$ (5)

so taking the Minus Sign and simplifying gives

$$\sin\left({ \pi\over 5}\right)= \sqrt{5-\sqrt{5}\over 8}={\textstyle{1\over 4}}\sqrt{10-2\sqrt{5}}.$$ (6)

$\cos(\pi/5)$ can be computed from

$$\cos\left({\pi\over 5}\right)= \sqrt{1-\sin^2\left({\pi\over 5}\right)} = {\textstyle{1\over 4}}(1+\sqrt{5}).$$ (7)

Summarizing,

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

$$\sin\left({2\pi\over 5}\right) = {\textstyle{1\over 4}}\sqrt{10+2\sqrt{5}}$$ (9)

$$\sin\left({3\pi\over 5}\right) = {\textstyle{1\over 4}}\sqrt{10+2\sqrt{5}}$$ (10)

$$\sin\left({4\pi\over 5}\right) = {\textstyle{1\over 4}}\sqrt{10-2\sqrt{5}}$$ (11)

$$\cos\left({ \pi\over 5}\right) = {\textstyle{1\over 4}}(1+\sqrt{5})$$ (12)

$$\cos\left({2\pi\over 5}\right) = {\textstyle{1\over 4}}(-1+\sqrt{5}\,)$$ (13)

$$\cos\left({3\pi\over 5}\right) = {\textstyle{1\over 4}}(1-\sqrt{5}\,)$$ (14)

$$\cos\left({4\pi\over 5}\right) = -{\textstyle{1\over 4}}(1+\sqrt{5}\,)$$ (15)

$$\tan\left({ \pi\over 5}\right) = \sqrt{5-2\sqrt{5}}$$ (16)

$$\tan\left({2\pi\over 5}\right) = \sqrt{5+2\sqrt{5}}$$ (17)

$$\tan\left({3\pi\over 5}\right) = -\sqrt{5+2\sqrt{5}}$$ (18)

$$\tan\left({4\pi\over 5}\right) = -\sqrt{5-2\sqrt{5}}\,.$$ (19)

See also Dodecahedron, Icosahedron, Pentagon, Pentagram