12.7 Special Values of Trigonometric Functions

We have

$$\cos\left({\pi\over 4}\right)=\cos\left({\pi\over 2}-{\pi\ove... ...{\pi\over 4}+\sin{\pi\over 2}\sin{\pi\over 4}=\sin{\pi\over 4}.$$

Hence $${1=\cos^2\left({\pi\over 4}\right)+\sin^2\left({\pi\over 4}\right)=2\sin^2\left({\pi\over 4}\right)}$$, and hence $${\left(\cos{\pi\over 4}\right) = \sin\left({\pi\over 4}\right)=\pm\sqrt{{1\over 2}}=\pm{{\sqrt 2}\over {2}}}$$. Since we know $\sin$ is positive on $(0,\pi)$, we conclude that

$$\cos\left({\pi\over 4}\right)=\sin\left({\pi\over 4}\right) ={{\sqrt 2}\over 2}.$$

Observe that if $t\in\mbox{{\bf R}}$, then the problem of calculating $\cos(t)$ and $\sin (t)$ is the same as the problem of calculating $e^{it}$. Let $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$. We know that the complex solutions of $z^n-1=0$ are

$$\left\{e^{{{2\pi ik}\over n}}\colon 0\leq k<n,\; k\in\mbox{{\bf Z}}\right\},$$

so if we can express the solutions to $z^n-1=0$ in algebraic terms, then we can express $${\sin\left( {{2\pi k}\over n}\right)}$$ and $${\cos\left( {{2\pi k}\over n}\right)}$$ in algebraic terms. We have

$$z^6-1=0\iff (z^3-1)(z^3+1)=0\iff (z-1)(z^2+z+1)(z+1)(z^2-z+1)=0.$$

Here $z=1$ and $z=-1$ are obvious sixth roots of $1$, and the other four roots are the solutions of the quadratic equations

$$z^2+z+1=0\mbox{ and }z^2-z+1=0.$$

12.67   Exercise. Find the solutions to $z^2+z+1=0$ and $z^2-z+1=0$ in terms of square roots of rational numbers. These solutions are

$$\left\{e^{ {{\pi i}\over 3}},e^{ {{2\pi i}\over 3}},e^{{{4\pi i}\over 3}},e^{{{5\pi i}\over 3}}\right\}.$$

Identify each solution with one of these exponentials. Find $${\cos\left({\pi\over 3}\right)}$$ and $${\sin\left({\pi\over 3}\right)}$$.

12.68   Exercise. Use the fact that

$$e^{{{\pi i}\over 6}}=e^{{{\pi i}\over 2}}\cdot e^{-{{\pi i}\over 3}}$$

to find $${\cos{\pi\over 6}}$$ and $${\sin{\pi\over 6}}$$. $\mid\!\mid\!\mid$

The numbers $${\cos \left({{2\pi}\over 5}\right)}$$ and $${\sin\left( {{2\pi}\over 5}\right)}$$ can also be expressed algebraically.

If $${z=e^{ {{2\pi i}\over 5}}}$$, then $z^5-1=0$, so

$$(z-1)(z^4+z^3+z^2+z+1)=0$$

and since $z\neq 1$,

$$(z^4+z^3+z^2+z+1)=0.$$

The fact that $z^5=1$ says $z^{-1}=z^4$ and $z^{-2}=z^3$, so

$$1+z+z^{-1}+z^2+z^{-2}=0;$$

i.e.,

$$1+e^{ {{2\pi i}\over 5}}+e^{ {{-2\pi i}\over 5}}+e^{ {{4\pi i}\over 5}}+ e^{ {{-4\pi i}\over 5}}=0,$$

or

$$1+2\cos\left({{2\pi}\over 5}\right)+2\cos\left({{4\pi}\over 5}\right)=0.$$

Now for all $z\in\mbox{{\bf C}}$,

$$\cos(2z)=\cos(z+z)=\cos^2(z)-\sin^2(z)=\cos^2(z)-\left(1-\cos^2(z)\right)=2\cos^2(z)-1,$$

so

$$1+2\cos\left({{2\pi}\over 5}\right)+2\left(2\cos^2\left({{2\pi}\over 5}\right)-1\right)=0.$$ (12.69)

Hence $${\cos \left({{2\pi}\over 5}\right)}$$ satisfies a quadratic equation.

12.70   Exercise. A $\quad$

a) Solve (12.69), and determine $${\cos \left({{2\pi}\over 5}\right)}$$ and $${\sin\left( {{2\pi}\over 5}\right)}$$ in algebraic terms.

b) The quadratic equation has two solutions, one of which is $${\cos \left({{2\pi}\over 5}\right)}$$. What is the geometrical significance of the other solution?

12.71   Entertainment. The algebraic representation for $${\cos \left({{2\pi}\over 5}\right)}$$ shows that a regular pentagon can be inscribed in a given circle. Let a circle be given, and call its radius $1$. If you can construct $${\cos\left({{2\pi}\over n}\right)}$$ with compass and straightedge (see the figure), then you can construct a side of a regular $n$-gon inscribed in the circle (and hence you can construct the $n$-gon).

For example, since $${\cos\left({{2\pi}\over {12}}\right)={{\sqrt 3}\over 2}}$$, we can construct a dodecagon as follows:

In the figure, make an arc of radius $2$ with center at $A$, intersecting the $x$-axis at $B$. Then $OB=\sqrt 3$, so if $C$ bisects $OB$, then $${OC=\cos\left({{2\pi}\over {12}}\right)}$$, and the vertical line through $C$ intersects the circle at $E$ where $IE$ is a side of the $12$-gon.

Use the formula for $${\cos \left({{2\pi}\over 5}\right)}$$ to inscribe a regular pentagon in a circle.

12.72   Entertainment. (This problem entertained Gauss. It will probably not really entertain you, unless you are another Gauss.) Show that a regular $17$-gon can be inscribed in a circle using compasses and straightedge.

Gauss discovered this result in 1796 [31, p 754] when he was a nineteen year old student at Göttingen. The result is [21, p 458]

$$\begin{eqnarray*} \cos\left({{2\pi}\over {17}}\right)&=&-{1\over {16}}+{1\over {... ...qrt{17+3\sqrt{17}-\sqrt{(34-2\sqrt{17})}-2\sqrt{34+2\sqrt{17}}}. \end{eqnarray*}$$