Trigonometry Values Pi/8

$$\sin\left({\pi\over 8}\right) = \sin\left({{1\over 2}\cdot{\pi\over 4}}\right) = \sqrt{{1\over 2}\left({1-\cos{\pi\over 4}}\right)}$$

$$= \sqrt{{\textstyle{1\over 2}}(1-{\textstyle{1\over 2}}\sqrt{2})} = {\textstyle{1\over 2}}\sqrt{2-\sqrt{2}}.$$ (1)

Now, checking to see if the Square Root can be simplified gives

$$a^2-b^2 c = 2^2-1^2\cdot 2 = 4-2 = 2,$$ (2)

which is not a Perfect Square, so the above expression cannot be simplified. Similarly,

$$\cos \left({\pi \over 8}\right) = \cos\left({{1 \over 2} {\pi \over 4}}\right)= \sqrt{{1\over 2}\left({1+\cos{\pi\over 4}}\right)}$$

$$= \sqrt{{1\over 2}\left({1+{\sqrt{2}\over 2}}\right)} = {\textstyle{1\over 2}}\sqrt{2+\sqrt{2}}$$ (3)

$$\tan\left({\pi \over 8}\right) = \sqrt{{2-\sqrt{2}}\over{2+\sqrt{2}}} = \sqrt{(2-\sqrt{2})^2 \over {4-2}} = \sqrt{{4+2-4\sqrt{2}} \over 2}$$

$$= \sqrt{{6-4\sqrt{2}} \over 2} = \sqrt{3-2\sqrt{2}}.$$ (4)

But

$$a^2-b^2 c = 3^2-2^2 2 = 9-8 = 1$$ (5)

is a Perfect Square, so we can find

$$d = {\textstyle{1\over 2}}(3\pm 1) = 1, 2.$$ (6)

Rewrite the above as

$$\tan\left({\pi \over 8}\right) = \sqrt{2}-1$$ (7)

$$\cot\left({\pi\over 8}\right) = {1 \over {\sqrt{2}-1}} = {{\sqrt{2}+1} \over {2-1}} = \sqrt{2}+1.$$ (8)

Summarizing,

$$\sin\left({\pi\over 8}\right) = {\textstyle{1\over 2}}\sqrt{2-\sqrt{2}}$$ (9)

$$\sin\left({3\pi\over 8}\right) = {\textstyle{1\over 2}}\sqrt{2+\sqrt{2}}$$ (10)

$$\cos \left({\pi \over 8}\right) = {\textstyle{1\over 2}}\sqrt{2+\sqrt{2}}$$ (11)

$$\cos\left({3\pi\over 8}\right) = {\textstyle{1\over 2}}\sqrt{2-\sqrt{2}}$$ (12)

$$\tan\left({\pi \over 8}\right) = \sqrt{2}-1$$ (13)

$$\tan\left({3\pi\over 8}\right) = \sqrt{2}+1.$$ (14)

See also Octagon