Trigonometry Values Pi/7

Trigonometric functions of $n\pi/7$ for $n$ an integer cannot be expressed in terms of sums, products, and finite root extractions on real rational numbers because 7 is not a Fermat Prime. This also means that the Heptagon is not a Constructible Polygon.

However, exact expressions involving roots of complex numbers can still be derived using the trigonometric identity

$$\sin(n\alpha)=2\sin[(n-1)\alpha]\cos\alpha-\sin[(n-2)\alpha].$$ (1)

The case $n=7$ gives

$\sin(7\alpha)=2\sin(6\alpha)\cos\alpha-\sin(5\alpha)$
$\quad =2(32\cos^5\alpha\sin\alpha-32\cos^3\alpha\sin\alpha+6\cos\alpha\sin\alpha)\cos\alpha$
$\quad \mathop{-}(5\sin\alpha-20\sin^3\alpha+16\sin^5\alpha)$
$= 64\cos^6\alpha\sin\alpha-64\cos^4\alpha\sin\alpha+12\cos^2\alpha\sin\alpha$
$\quad -5\sin\alpha+20(1-\cos^2\alpha)\sin\alpha$
$\quad -16(1-2\cos^2\alpha+\cos^4\alpha)\sin\alpha$
$\quad =\sin\alpha(64\cos^6\alpha-80\cos^4\alpha+24\cos^2\alpha-1).$	(2)

Rewrite this using the identity $\cos^2\alpha = 1-\sin^2\alpha$,

$\sin\left({\pi\over7}\right)=\sin\alpha(7-56\sin^2\alpha+112\sin^4\alpha-64\sin^6\alpha)$
$=-64\sin\alpha(\sin^6\alpha-{\textstyle{112\over 64}}\sin^4\alpha+{\textstyle{56\over 64}}\sin^2\alpha-{\textstyle{7\over 64}}).\quad$	(3)

Now, let $\alpha \equiv \pi/7$ and $x \equiv \sin^2\alpha$, then

$$\sin(\pi) = 0 = x^3-{\textstyle{7\over 4}} x^2+{\textstyle{7\over 8}} x - {\textstyle{7\over 64}},$$ (4)

which is a Cubic Equation in $x$. The Roots are numerically found to be $x\approx 0.188255$, $0.611260\ldots$, $0.950484\ldots$. But $\sin \alpha = \sqrt{x}$, so these Roots correspond to $\sin\alpha\approx 0.4338$, $\sin(2\alpha)\approx 0.7817$, $\sin(3\alpha)\approx 0.9749$. By Newton's Relation

$$\prod_i r_i = -a_0,$$ (5)

we have

$$x_1 x_2 x_3 = {\textstyle{7\over 64}},$$ (6)

or

$$\sin\left({\pi\over 7}\right)\sin\left({2\pi\over 7}\right)\... ... 7}\right)= \sqrt{7\over 64} = {\textstyle{1\over 8}}\sqrt{7}.$$ (7)

Similarly,

$$\cos\left({\pi\over 7}\right)\cos\left({2\pi\over 7}\right)\cos\left({3\pi\over 7}\right)= {1\over 8}.$$ (8)

The constants of the Cubic Equation are given by

$$Q \equiv {\textstyle{1\over 9}} (3 a_1 - {a_2}^2) = {\textstyle{1\over 9}}... ...\textstyle{7\over 8}}-(-{\textstyle{7\over 4}})^2] = - {\textstyle{7\over 144}}$$ (9)

$$R \equiv {\textstyle{1\over 54}} (9 a_2 a_1-2a_2^3-27 a_0)$$

$$= {\textstyle{1\over 54}}[9(-{\textstyle{7\over 4}})({\textstyle{1\over 7}}{8})-2(-{\textstyle{7\over 4}})^3-27(-{\textstyle{7\over 64}})]$$

$$= - {\textstyle{7\over 3456}}.$$ (10)

The Discriminant is then

$$D \equiv Q^3+R^2 = - {\textstyle{343\over 2,985,984}}+{\textstyle{49\over 11,943,936}}$$

$$= - {\textstyle{49\over 442,368}} < 0,$$ (11)

so there are three distinct Real Roots. Finding the first one,

$$x = {\root 3 \of {R+\sqrt{D}}}+{\root 3 \of {R-\sqrt{D}}}-{\textstyle{1\over 3}} a_2.$$ (12)

Writing

$$\sqrt{D} = 3^{-3/2} {\textstyle{7\over 128}} i,$$ (13)

plugging in from above, and anticipating that the solution we have picked corresponds to $\sin(3\pi/7)$,

$\sin\left({3\pi \over 7}\right)= \sqrt{x}=\sqrt{{\root 3 \of{-{7\over 3456}+3^{... ...{-{7\over 3456}-3^{-3/2} {7\over 128} i}}-{1\over 3}\left({-{7\over 4}}\right)}$
$\quad = \sqrt{{\root 3 \of{-{7\over 3456}+3^{-3/2} {7\over 128} i}}+{\root 3\of{-{7\over 3456}-3^{-3/2} {7\over 128} i}}+{7\over 12}}$
$\quad = \sqrt{{\root 3 \of{{7\over 3456}\left({-1+3^{3/2} i}\right)}}- {\root 3 \of{{7\over 3456}\left({1+3^{3/2} i}\right)}}+{7\over 12}}$
$\quad = \sqrt{{1\over 12}\left[{{\root 3 \of{{7\over 2}\left({-1+3^{3/2} i}\right)}}- {\root 3\of{{7\over 2}\left({1+3^{3/2} i}\right)}}+7}\right]}.$	(14)

See also Heptagon