Calculus – Trigonometry – Special values

Using the standard values of the sine, cosine and tangent together with the summation formulas, we are able to derive exact special values. For instance, since \(\frac{1}{12}\pi=\frac{1}{3}\pi-\frac{1}{4}\pi\), we obtain

\[\sin(\tfrac{1}{12}\pi)=\sin(\tfrac{1}{3}\pi-\tfrac{1}{4}\pi)=\sin(\tfrac{1}{3}\pi)\cos(\tfrac{1}{4}\pi)-\cos(\tfrac{1}{3}\pi)\sin(\tfrac{1}{4}\pi) =\tfrac{1}{2}\sqrt{3}\cdot\tfrac{1}{2}\sqrt{2}-\tfrac{1}{2}\cdot\tfrac{1}{2}\sqrt{2}=\tfrac{1}{4}\sqrt{6}-\tfrac{1}{4}\sqrt{2}\]

and

\[\cos(\tfrac{1}{12}\pi)=\cos(\tfrac{1}{3}\pi-\tfrac{1}{4}\pi)=\cos(\tfrac{1}{3}\pi)\cos(\tfrac{1}{4}\pi)+\sin(\tfrac{1}{3}\pi)\sin(\tfrac{1}{4}\pi) =\tfrac{1}{2}\cdot\tfrac{1}{2}\sqrt{2}+\tfrac{1}{2}\sqrt{3}\cdot\tfrac{1}{2}\sqrt{2}=\tfrac{1}{4}\sqrt{6}+\tfrac{1}{4}\sqrt{2}.\]

Similarly, using \(\frac{5}{12}\pi=\frac{1}{6}\pi+\frac{1}{4}\pi\), we obtain

\[\sin(\tfrac{5}{12}\pi)=\sin(\tfrac{1}{6}\pi+\tfrac{1}{4}\pi)=\sin(\tfrac{1}{6}\pi)\cos(\tfrac{1}{4}\pi)+\cos(\tfrac{1}{6}\pi)\sin(\tfrac{1}{4}\pi) =\tfrac{1}{2}\cdot\tfrac{1}{2}\sqrt{2}+\tfrac{1}{2}\sqrt{3}\cdot\tfrac{1}{2}\sqrt{2}=\tfrac{1}{4}\sqrt{6}+\tfrac{1}{4}\sqrt{2}\]

and

\[\cos(\tfrac{5}{12}\pi)=\cos(\tfrac{1}{6}\pi+\tfrac{1}{4}\pi)=\cos(\tfrac{1}{6}\pi)\cos(\tfrac{1}{4}\pi)-\sin(\tfrac{1}{6}\pi)\sin(\tfrac{1}{4}\pi) =\tfrac{1}{2}\sqrt{3}\cdot\tfrac{1}{2}\sqrt{2}-\tfrac{1}{2}\cdot\tfrac{1}{2}\sqrt{2}=\tfrac{1}{4}\sqrt{6}-\tfrac{1}{4}\sqrt{2}.\]

Furthermore, we have

\[\sin(\tfrac{7}{12}\pi)=\sin(\pi-\tfrac{5}{12}\pi)=\sin(\tfrac{5}{12}\pi)=\tfrac{1}{4}\sqrt{6}+\tfrac{1}{4}\sqrt{2}\quad\text{and}\quad \cos(\tfrac{7}{12}\pi)=\cos(\pi-\tfrac{5}{12}\pi)=-\cos(\tfrac{5}{12}\pi)=\tfrac{1}{4}\sqrt{2}-\tfrac{1}{4}\sqrt{6}.\]

Similarly, we have

\[\sin(\tfrac{11}{12}\pi)=\sin(\pi-\tfrac{1}{12}\pi)=\sin(\tfrac{1}{12}\pi)=\tfrac{1}{4}\sqrt{6}-\tfrac{1}{4}\sqrt{2}\quad\text{and}\quad \cos(\tfrac{11}{12}\pi)=\cos(\pi-\tfrac{1}{12}\pi)=-\cos(\tfrac{1}{12}\pi)=-\tfrac{1}{4}\sqrt{6}-\tfrac{1}{4}\sqrt{2}.\]

For the tangent we then obtain:

\[\tan(\tfrac{1}{12}\pi)=\frac{\sin(\frac{1}{12}\pi)}{\cos(\frac{1}{12}\pi)}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}, \quad\tan(\tfrac{5}{12}\pi)=\frac{\sin(\frac{5}{12}\pi)}{\cos(\frac{5}{12}\pi)}=\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{3}+1}{\sqrt{3}-1},\] \[\tan(\tfrac{7}{12}\pi)=\frac{\sin(\frac{7}{12}\pi)}{\cos(\frac{7}{12}\pi)}=\frac{\sqrt{2}+\sqrt{6}}{\sqrt{2}-\sqrt{6}}=\frac{1+\sqrt{3}}{1-\sqrt{3}}, \quad\tan(\tfrac{11}{12}\pi)=\frac{\sin(\frac{11}{12}\pi)}{\cos(\frac{11}{12}\pi)}=\frac{\sqrt{2}-\sqrt{6}}{\sqrt{2}+\sqrt{6}}=\frac{1-\sqrt{3}}{1+\sqrt{3}}.\]

Let \(a=\cos(\frac{1}{5}\pi)\) and \(b=\cos(\tfrac{2}{5}\pi)\). Now we use \(\cos(2x)=2\cos^2(x)-1\) and \(\cos(\pi-x)=-\cos(x)\) to find that

\[\cos(\tfrac{2}{5}\pi)=2\cos^2(\tfrac{1}{5}\pi)-1\quad\Longleftrightarrow\quad b=2a^2-1\]

and

\[-\cos(\tfrac{1}{5}\pi)=\cos(\tfrac{4}{5}\pi)=2\cos^2(\tfrac{2}{5}\pi)-1\quad\Longleftrightarrow\quad -a=2b^2-1.\]

Subtracting we obtain \(a+b=2a^2-2b^2=2(a+b)(a-b)\), which implies that \(1=2(a-b)\) since \(a+b\neq0\). Substituting \(b=a-\frac{1}{2}\) we obtain

\[a-\tfrac{1}{2}=2a^2-1\quad\Longleftrightarrow\quad 4a^2-2a-1=0\quad\Longleftrightarrow\quad(2a-\tfrac{1}{2})^2=\tfrac{5}{4} \quad\Longleftrightarrow\quad a=\tfrac{1}{4}(1\pm\sqrt{5}).\]

Since \(a=\cos(\frac{1}{5}\pi)>0\), we conclude that \(a=\frac{1}{4}(1+\sqrt{5})\). This implies that \(b=a-\frac{1}{2}=\frac{1}{4}(\sqrt{5}-1)\). We conclude that

\[\cos(\tfrac{1}{5}\pi)=\tfrac{1}{4}(1+\sqrt{5}),\quad\cos(\tfrac{2}{5}\pi)=\tfrac{1}{4}(\sqrt{5}-1),\quad \cos(\tfrac{3}{5}\pi)=-\cos(\tfrac{2}{5}\pi)=\tfrac{1}{4}(1-\sqrt{5}),\quad\cos(\tfrac{4}{5}\pi)=-\cos(\tfrac{1}{5}\pi)=-\tfrac{1}{4}(1+\sqrt{5}).\]

Using \(\cos(2x)=1-2\sin^2(x)\), we now have:

\[\cos(\tfrac{1}{5}\pi)=1-2\sin^2(\tfrac{1}{10}\pi)\quad\Longleftrightarrow\quad2\sin^2(\tfrac{1}{10}\pi)=1-\tfrac{1}{4}(1+\sqrt{5})=\tfrac{1}{4}(3-\sqrt{5}) \quad\Longleftrightarrow\quad\sin^2(\tfrac{1}{10}\pi)=\frac{3-\sqrt{5}}{8}=\left(\frac{\sqrt{5}-1}{4}\right)^2.\]

Since \(\sin(\frac{1}{10}\pi)>0\), we conclude that \(\sin(\frac{1}{10}\pi)=\frac{1}{4}(\sqrt{5}-1)\). Another derivation is: \(\sin(\frac{1}{10}\pi)=-\cos(\frac{1}{10}\pi+\frac{1}{2}\pi)=-\cos(\frac{3}{5}\pi)\).

Similarly, we have: \(\sin(\frac{3}{10}\pi)=-\cos(\frac{3}{10}\pi+\frac{1}{2}\pi)=-\cos(\frac{4}{5}\pi)=\cos(\frac{1}{5}\pi)=\frac{1}{4}(1+\sqrt{5})\).

Using the double angle formula \(\tan(2x)=\dfrac{2\tan(x)}{1-\tan^2(x)}\) we obtain for \(x=\frac{1}{8}\pi\):

\[1=\tan(\tfrac{1}{4}\pi)=\frac{2\tan(\frac{1}{8}\pi)}{1-\tan^2(\frac{1}{8}\pi)}\quad\Longleftrightarrow\quad 1-\tan^2(\tfrac{1}{8}\pi)=2\tan(\tfrac{1}{8}\pi) \quad\Longleftrightarrow\quad\tan^2(\tfrac{1}{8}\pi)+2\tan(\tfrac{1}{8}\pi)-1=0.\]

Hence: \(\left(\tan(\tfrac{1}{8}\pi)+1\right)^2-2=0\) or equivalently \(\tan(\tfrac{1}{8}\pi)=-1\pm\sqrt{2}\). Since \(\tan(\tfrac{1}{8}\pi)>0\) we conclude that \(\tan(\tfrac{1}{8}\pi)=-1+\sqrt{2}\).

Similarly we obtain for \(x=\frac{3}{8}\pi\):

\[-1=\tan(\tfrac{3}{4}\pi)=\frac{2\tan(\frac{3}{8}\pi)}{1-\tan^2(\frac{3}{8}\pi)}\quad\Longleftrightarrow\quad -1+\tan^2(\tfrac{3}{8}\pi)=2\tan(\tfrac{3}{8}\pi) \quad\Longleftrightarrow\quad\tan^2(\tfrac{3}{8}\pi)-2\tan(\tfrac{3}{8}\pi)-1=0.\]

Hence: \(\left(\tan(\tfrac{3}{8}\pi)-1\right)^2-2=0\) or equivalently \(\tan(\tfrac{3}{8}\pi)=1\pm\sqrt{2}\). Since \(\tan(\tfrac{3}{8}\pi)>0\) we conclude that \(\tan(\tfrac{3}{8}\pi)=1+\sqrt{2}\).

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Last modified on August 6, 2022