Exact Trigonometric Values

Exact trigonometric values are those values of $$\cos x,\ \sin x$$ and $$\tan x$$ that can be calculated exactly.

The calculations of these values vary from very simple to somewhat complicated.

Imagine a right triangle in which base and hypotenuse are equal and height is $$0.$$ Then

Imagine a right triangle in which height and hypotenuse are equal and base is $$0.$$ Then

From the right isosceles triangle:

From the equilateral triangle:

For derivation of $$\sin 18^\circ,$$ see Sine of 18°.

The derivation of $$\sin 18^\circ$$ also produces the value $$\frac{-\sqrt{5} - 1}{4}$$ which is in fact $$\sin -54^\circ.$$ therefore:

=Derivation of cos 18°=

Let $$\cos 18^\circ = x.$$ Then:

$$\sin 54^\circ = \cos 36^\circ = \cos(2 \cdot 18^\circ) = 2x^2 - 1.$$

$$\cos 54^\circ = \cos(3 \cdot 18^\circ) = 4x^3 - 3x.$$

$$\cos^2 54^\circ + \sin^2 54^\circ = 1.$$ Therefore:

$$(4x^3 - 3x)^2 + (2x^2 - 1)^2 = 1.$$

Simplify and result is:

$$16x^4 - 20x^2 + 5 = 0$$ or:

$$16X^2 - 20X + 5 = 0$$ where $$X = x^2$$ or $$x = \sqrt{X}.$$

$$X = \frac{5 \pm \sqrt{5}}{8}$$

$$\sin 72^\circ = \cos 18^\circ = \sqrt{ \frac{5 + \sqrt{5}}{8} }$$

$$\sin 36^\circ = \cos 54^\circ = \sqrt{ \frac{5 - \sqrt{5}}{8} }$$

=Derivation of cos 36°=

Let $$\cos 36^\circ = x.$$

Then $$\cos 72^\circ = \cos (2\cdot 36^\circ) = 2x^2 - 1$$ and

$$\cos 108^\circ = \cos(3\cdot 36^\circ) = 4x^3 - 3x.$$

$$\cos 108^\circ = -\cos 72^\circ.$$

Therefore:

$$4x^3 - 3x = -(2x^2 - 1)$$ or

$$4x^3 + 2x^2  - 3x - 1 = 0.$$

This cubic equation contains the root $$x = \cos 180^\circ = -1.$$

Remove factor $$(x+1)$$ and remaining quadratic is:

$$4x^2 - 2x -1 = 0.$$

Solutions of this quadratic are: $$\frac{1 \pm \sqrt{5}}{4}.$$

$$\cos 36^\circ = \frac{1 + \sqrt{5}}{4}.$$

$$\cos 108^\circ = \frac{1 - \sqrt{5}}{4}.$$

$$\sin 18^\circ = \cos 72^\circ = -\cos 108^\circ = \frac{-1 + \sqrt{5}}{4}.$$

=Calculation of tan 18°=

$$\tan 18^\circ = \frac{\sin 18^\circ}{\cos 18^\circ}$$ $$= \frac{\sqrt{5}-1}{4}\cdot\sqrt{\frac{8}{\sqrt{5}+5}}$$

$$\tan^2 18^\circ = \frac{\sqrt{5}-1}{4}\cdot\frac{\sqrt{5}-1}{4}\cdot\frac{8}{\sqrt{5}+5}$$ $$= \frac{5-2\sqrt{5}}{5}$$

$$\tan 18^\circ = \sqrt{ \frac{5-2\sqrt{5}}{5}   }$$ $$= \sqrt{ \frac{5\cdot 5-5\cdot 2\sqrt{5}}{5\cdot 5}   }$$ $$= \frac{\sqrt{25 - 10\sqrt{5}}}{5}$$

=Values for 18° and 36°=

=Calculation of cos 9°=

Using half-angle formula
$$\cos^2 9^\circ = \frac{1 + \cos 18^\circ}{2}$$ $$= \frac{1 + \sqrt{\frac{\sqrt{5}+5}{8}}}{2}$$ $$= \frac{1 + \sqrt{\frac{2\sqrt{5}+10}{16}}}{2}$$ $$= \frac{1 + \frac{\sqrt{2\sqrt{5}+10}}{4}}{2}$$ $$= \frac{4 + \sqrt{2\sqrt{5}+10}}{8}$$

$$\cos 9^\circ = \sqrt{ \frac{4 + \sqrt{2\sqrt{5}+10}}{8} }$$

Using difference formula

 * {| class="wikitable" style="text-align: center;"

!Radians!!Degrees!!$sin$!!$cos$ ! $\pi⁄5$!! $36°$ ! $\pi⁄4$!! $45°$
 * $$\frac{\sqrt{10- 2\sqrt{5}}} {4}$$||$$\frac{1+ \sqrt{5}} {4}$$
 * $$\frac{\sqrt{2}}{2}$$|| $$\frac{\sqrt{2}}{2}$$
 * }

$$\cos(A-B) = \cos A \cdot \cos B + \sin A \cdot \sin B$$

$$\cos 9^\circ = \cos(45^\circ - 36^\circ)$$ $$= \cos 45^\circ \cdot \cos 36^\circ + \sin 45^\circ \cdot \sin 36^\circ$$ $$= \frac{\sqrt{2}}{2} \cdot \frac{1+ \sqrt{5}} {4} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{10- 2\sqrt{5}}} {4}$$ $$= \frac{\sqrt{2}(1+\sqrt{5})}{8}    + \frac {\sqrt{5 - \sqrt{5}}}{4}     $$

Although the two calculated values of $$\cos 9^\circ$$ are expressed differently, it can be shown that the two values $$\sqrt{ \frac{4 + \sqrt{2\sqrt{5}+10}}{8} }$$ and $$\frac{\sqrt{2}(1+\sqrt{5})}{8}    + \frac {\sqrt{5 - \sqrt{5}}}{4}     $$ are equal.

By using similar techniques many exact values can be calculated.

=Exact Values for Common Angles=


 * {| class="wikitable" style="text-align: center;"

!Radians!!Degrees!!$sin$!!$cos$!!$tan$ !$$ 0$$!! $0°$ ! $\pi⁄60$!! $3°$   || $$\frac{ \sqrt{8 + \sqrt{3}( \sqrt{5}+1 )  + \sqrt{10-2\sqrt{5}}  }      }{4} $$ || $$\sqrt{\frac{8-\sqrt{3}(\sqrt{5}+1)-\sqrt{10-2\sqrt{5}}} {8+\sqrt{3}(\sqrt{5}+1)+\sqrt{10-2\sqrt{5}}}}$$ ! $\pi⁄30$!! $6°$   || $$\frac{ \sqrt{3}( \sqrt{5}+1 )  + \sqrt{10-2\sqrt{5}}        }{8}$$ || $$\frac{\sqrt{5-2\sqrt{5}} \cdot (\sqrt{5}+1) + \sqrt{3} \cdot (1-\sqrt{5})}   {2}$$ ! $\pi⁄20$!! $9°$   || $$\sqrt{\frac{4 + \sqrt{10+2\sqrt{5}}} {8}}$$ || $$\sqrt{ 11 + 4\sqrt{5} - (3+\sqrt{5})\sqrt{10 + 2\sqrt{5}}  }$$ ! $\pi⁄15$!! $12°$   || $$\frac{  \sqrt{9 + \sqrt{5} + \sqrt{30-6\sqrt{5}}}    }{4}$$ || $$\frac{ \sqrt{ 92 - 40\sqrt{5} + (6\sqrt{5} - 14)\sqrt{30 - 6\sqrt{5}} } } {2}$$ ! $\pi⁄12$!! $15°$ ! $\pi⁄10$!! $18°$ ! $7\pi⁄60$!! $21°$             }$$             }$$                   {4 + \sqrt{7 + \sqrt{30-6\sqrt{5}}  - \sqrt{5}}} }$$ ! $\pi⁄8$|| $22.5°$ ! $2\pi⁄15$!! $24°$   || $$\frac{1 + \sqrt{5} + \sqrt{30-6\sqrt{5}}}{8}$$ || $$\frac{(3\sqrt{5}+7)\sqrt{5 - 2\sqrt{5}} - (\sqrt{5} + 3)\sqrt{3} }{2}$$ ! $3\pi⁄20$!! $27°$ ! $\pi⁄6$!! $30°$ ! $11\pi⁄60$!! $33°$    || $$\frac{    \sqrt{ 8 + \sqrt{15} + \sqrt{3} - \sqrt{10 - 2\sqrt{5}} }  }{4}$$ || $$\sqrt{\frac{8 - \sqrt{15} - \sqrt{3} + \sqrt{10 - 2\sqrt{5}}}{8 + \sqrt{15} + \sqrt{3} - \sqrt{10 - 2\sqrt{5}}}}$$ ! $\pi⁄5$!! $36°$ ! $13\pi⁄60$!! $39°$   \sqrt{  8 - 2\sqrt{   7 - \sqrt{5} - \sqrt{30-6\sqrt{5}} }    } }{4}$$    ||$$\frac{ \sqrt{ 8 + 2\sqrt{   7 - \sqrt{5} - \sqrt{30-6\sqrt{5}} }    } }{4}$$    ||$$\sqrt{\frac{  8 - 2\sqrt{   7 - \sqrt{5} - \sqrt{30-6\sqrt{5}} }    }{  8         + 2\sqrt{    7 - \sqrt{5} - \sqrt{30-6\sqrt{5}} }    }}$$ ! $7\pi⁄30$!! $42°$                   {  7 + \sqrt{30-6\sqrt{5}}  - \sqrt{5}      } }$$ ! $\pi⁄4$!! $45°$
 * $$ 0$$||$$1$$||$$ 0$$
 * $$\frac{ \sqrt{8 - \sqrt{3}( \sqrt{5}+1 ) - \sqrt{10-2\sqrt{5}}  }      }{4} $$
 * $$\frac{\sqrt{30-6\sqrt{5}} - \sqrt{5}-1}{8}$$
 * $$\sqrt{\frac{4 - \sqrt{10+2\sqrt{5}}} {8}}$$
 * $$\frac{\sqrt{   7 - \sqrt{5} - \sqrt{30-6\sqrt{5}}     }}{4}$$
 * $$\frac{ \sqrt{6} - \sqrt{2} } {4}$$|| $$\frac{\sqrt{6}+\sqrt{2}}{4}$$|| $$2-\sqrt{3}$$
 * $$\frac{\sqrt{5}-1}{4}$$
 * $$\frac{\sqrt{10+2\sqrt{5}}} {4}$$
 * $$\frac{\sqrt{25-10\sqrt5}} {5}$$
 * $$\sqrt{\frac{4 - \sqrt{7 + \sqrt{30-6\sqrt{5}} - \sqrt{5}}}{8}
 * $$\sqrt{\frac{4 + \sqrt{7 + \sqrt{30-6\sqrt{5}} - \sqrt{5}}}{8}
 * $$\sqrt{\frac{4 - \sqrt{7 + \sqrt{30-6\sqrt{5}} - \sqrt{5}}}
 * $$\frac{ \sqrt{2 - \sqrt{2}} } {2}$$|| $$\frac{ \sqrt{2 + \sqrt{2}} } {2}$$|| $$\sqrt{2} - 1$$
 * $$\frac{\sqrt{15} + \sqrt{3} - \sqrt{10 - 2\sqrt{5}}}{8}$$
 * $$\sqrt{\frac{4 - \sqrt{10- 2\sqrt{5}}} {8}}$$
 * $$\sqrt{\frac{4 + \sqrt{10- 2\sqrt{5}}} {8}}$$
 * $$\sqrt{ 11 - 4\sqrt{5} +(\sqrt{5} - 3)\sqrt{10-2\sqrt{5}} }$$
 * $$\frac{1}{2}$$|| $$\frac{\sqrt{3}}{2}$$|| $$\frac{\sqrt{3}}{3}$$
 * $$\frac{   \sqrt{ 8 - \sqrt{15} - \sqrt{3} + \sqrt{10 - 2\sqrt{5}} }  }{4}$$
 * $$\frac{\sqrt{10- 2\sqrt{5}}} {4}$$||$$\frac{1+ \sqrt{5}} {4}$$||$$\sqrt{5-2\sqrt5}$$
 * $$\frac{
 * $$\frac{\sqrt{9 - \sqrt{30-6\sqrt{5}} + \sqrt{5}}}{4}  $$
 * $$\frac{\sqrt{7 + \sqrt{30-6\sqrt{5}} - \sqrt{5}}}{4}$$
 * $$\sqrt{\frac{ 9 - \sqrt{30-6\sqrt{5}}  + \sqrt{5}      }
 * $$\frac{\sqrt{2}}{2}$$|| $$\frac{\sqrt{2}}{2}$$|| $$1$$
 * }

Other Expressions of Exact Values
Depending on how the value is calculated, exact trigonometric values can be expressed in different ways. For example:

$$\tan 9^\circ = 1 + \sqrt{5} - (\sqrt{5} + 2)\sqrt{(5 - 2\sqrt{5})}$$

This value of $$\tan 9^\circ$$ contains calculation of 2 square roots. Value shown in table above contains calculation of 3 square roots.

$$\tan 27^\circ = \sqrt{5} - 1 - \sqrt{5 - 2\sqrt{5}}$$

$$\tan 33^\circ = \frac{ ( 2 - (2 - \sqrt{3})(3 + \sqrt{5}) )\ \cdot\ (2 + \sqrt{2(5 - \sqrt{5})})   }{4}$$

This value of $$\tan 33^\circ$$ contains calculation of 3 square roots. Value shown in table above contains calculation of 4 square roots.

Create a Dictionary
If you would like to create a dictionary containing the above values, the following python code has been tested on a Mac:

Check the Dictionary
The following python code is used to check the contents of the dictionary and to verify that there are no obvious errors.

=Links to related Topics=

Angle sum and difference identities

Angle sum identities

Double-angle formulae

Half-angle formulae

Simple algebraic values

Common angles

Using fifth roots of unity to calculate Cosine of 72°.

Calculating π

Testing taylor series for $$\arctan x$$