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12.7 Special Values of Trigonometric Functions

We have

\begin{displaymath}\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}.\end{displaymath}

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

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

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

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

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

\begin{displaymath}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.\end{displaymath}

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

\begin{displaymath}z^2+z+1=0\mbox{ and }z^2-z+1=0.\end{displaymath}

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

\begin{displaymath}\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\}.\end{displaymath}

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

12.68   Exercise. Use the fact that

\begin{displaymath}e^{{{\pi i}\over 6}}=e^{{{\pi i}\over 2}}\cdot e^{-{{\pi i}\over 3}}\end{displaymath}

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

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

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


and since $z\neq 1$,


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



\begin{displaymath}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,\end{displaymath}


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

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


1+2\cos\left({{2\pi}\over 5}\right)+2\left(2\cos^2\left({{2\pi}\over
\end{displaymath} (12.69)

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

12.70   Exercise. A $\quad$
a) Solve (12.69), and determine $\displaystyle {\cos \left({{2\pi}\over 5}\right)}$ and $\displaystyle {\sin\left( {{2\pi}\over 5}\right)}$ in algebraic terms.
b) The quadratic equation has two solutions, one of which is $\displaystyle {\cos \left({{2\pi}\over 5}\right)}$. What is the geometrical significance of the other solution?

12.71   Entertainment. The algebraic representation for $\displaystyle {\cos \left({{2\pi}\over 5}\right)}$ 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 $\displaystyle {\cos\left({{2\pi}\over n}\right)}$ 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 $\displaystyle {\cos\left({{2\pi}\over {12}}\right)={{\sqrt 3}\over 2}}$, 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 $\displaystyle {OC=\cos\left({{2\pi}\over {12}}\right)}$, and the vertical line through $C$ intersects the circle at $E$ where $IE$ is a side of the $12$-gon.

Use the formula for $\displaystyle {\cos \left({{2\pi}\over 5}\right)}$ 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]

\cos\left({{2\pi}\over {17}}\right)&=&-{1\over {16}}+{1\over {...

next up previous index
Next: 12.8 Proof of the Up: 12. Power Series Previous: 12.6 Trigonometric Functions   Index