# Exact Trigonometric Values

Exact trigonometric values are those values of ${\displaystyle \cos x,\ \sin x}$ and ${\displaystyle \tan x}$ 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 ${\displaystyle 0.}$ Then

x in radians x in degrees ${\displaystyle \sin(x)}$ ${\displaystyle \cos(x)}$ ${\displaystyle \tan(x)}$
${\displaystyle 0}$ ${\displaystyle 0^{\circ }}$ ${\displaystyle 0}$ ${\displaystyle 1}$ ${\displaystyle 0}$

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

x in radians x in degrees ${\displaystyle \sin(x)}$ ${\displaystyle \cos(x)}$ ${\displaystyle \tan(x)}$
${\displaystyle {\frac {\pi }{2}}}$ ${\displaystyle 90^{\circ }}$ ${\displaystyle 1}$ ${\displaystyle 0}$ ${\displaystyle {\frac {1}{0}}}$

From the right isosceles triangle:

x in radians x in degrees ${\displaystyle \sin(x)}$ ${\displaystyle \cos(x)}$ ${\displaystyle \tan(x)}$
${\displaystyle {\frac {\pi }{4}}}$ ${\displaystyle 45^{\circ }}$ ${\displaystyle {\frac {\sqrt {2}}{2}}}$ ${\displaystyle {\frac {\sqrt {2}}{2}}}$ ${\displaystyle 1}$

From the equilateral triangle:

x in radians x in degrees ${\displaystyle \sin(x)}$ ${\displaystyle \cos(x)}$ ${\displaystyle \tan(x)}$
${\displaystyle {\frac {\pi }{6}}}$ ${\displaystyle 30^{\circ }}$ ${\displaystyle {\frac {1}{2}}}$ ${\displaystyle {\frac {\sqrt {3}}{2}}}$ ${\displaystyle {\frac {\sqrt {3}}{3}}}$
${\displaystyle {\frac {\pi }{3}}}$ ${\displaystyle 60^{\circ }}$ ${\displaystyle {\frac {\sqrt {3}}{2}}}$ ${\displaystyle {\frac {1}{2}}}$ ${\displaystyle {\sqrt {3}}}$

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

x in radians x in degrees ${\displaystyle \sin(x)}$ ${\displaystyle \cos(x)}$ ${\displaystyle \tan(x)}$
${\displaystyle {\frac {\pi }{10}}}$ ${\displaystyle 18^{\circ }}$ ${\displaystyle {\frac {{\sqrt {5}}-1}{4}}}$ ${\displaystyle \dots }$ ${\displaystyle \dots }$
${\displaystyle {\frac {2\pi }{5}}}$ ${\displaystyle 72^{\circ }}$ ${\displaystyle \dots }$ ${\displaystyle {\frac {{\sqrt {5}}-1}{4}}}$ ${\displaystyle \dots }$

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

x in radians x in degrees ${\displaystyle \sin(x)}$ ${\displaystyle \cos(x)}$ ${\displaystyle \tan(x)}$
${\displaystyle {\frac {\pi }{5}}}$ ${\displaystyle 36^{\circ }}$ ${\displaystyle \dots }$ ${\displaystyle {\frac {{\sqrt {5}}+1}{4}}}$ ${\displaystyle \dots }$
${\displaystyle {\frac {3\pi }{10}}}$ ${\displaystyle 54^{\circ }}$ ${\displaystyle {\frac {{\sqrt {5}}+1}{4}}}$ ${\displaystyle \dots }$ ${\displaystyle \dots }$

# Derivation of cos 18°

 Let ${\displaystyle \cos 18^{\circ }=x.}$ Then: ${\displaystyle \sin 54^{\circ }=\cos 36^{\circ }=\cos(2\cdot 18^{\circ })=2x^{2}-1.}$ ${\displaystyle \cos 54^{\circ }=\cos(3\cdot 18^{\circ })=4x^{3}-3x.}$ ${\displaystyle \cos ^{2}54^{\circ }+\sin ^{2}54^{\circ }=1.}$ Therefore: ${\displaystyle (4x^{3}-3x)^{2}+(2x^{2}-1)^{2}=1.}$ Simplify and result is: ${\displaystyle 16x^{4}-20x^{2}+5=0}$ or: ${\displaystyle 16X^{2}-20X+5=0}$ where ${\displaystyle X=x^{2}}$ or ${\displaystyle x={\sqrt {X}}.}$ ${\displaystyle X={\frac {5\pm {\sqrt {5}}}{8}}}$ ${\displaystyle \sin 72^{\circ }=\cos 18^{\circ }={\sqrt {\frac {5+{\sqrt {5}}}{8}}}}$ ${\displaystyle \sin 36^{\circ }=\cos 54^{\circ }={\sqrt {\frac {5-{\sqrt {5}}}{8}}}}$

# Derivation of cos 36°

 Let ${\displaystyle \cos 36^{\circ }=x.}$ Then ${\displaystyle \cos 72^{\circ }=\cos(2\cdot 36^{\circ })=2x^{2}-1}$ and ${\displaystyle \cos 108^{\circ }=\cos(3\cdot 36^{\circ })=4x^{3}-3x.}$ ${\displaystyle \cos 108^{\circ }=-\cos 72^{\circ }.}$ Therefore: ${\displaystyle 4x^{3}-3x=-(2x^{2}-1)}$ or ${\displaystyle 4x^{3}+2x^{2}-3x-1=0.}$ This cubic equation contains the root ${\displaystyle x=\cos 180^{\circ }=-1.}$ Remove factor ${\displaystyle (x+1)}$ and remaining quadratic is: ${\displaystyle 4x^{2}-2x-1=0.}$ Solutions of this quadratic are: ${\displaystyle {\frac {1\pm {\sqrt {5}}}{4}}.}$ ${\displaystyle \cos 36^{\circ }={\frac {1+{\sqrt {5}}}{4}}.}$ ${\displaystyle \cos 108^{\circ }={\frac {1-{\sqrt {5}}}{4}}.}$ ${\displaystyle \sin 18^{\circ }=\cos 72^{\circ }=-\cos 108^{\circ }={\frac {-1+{\sqrt {5}}}{4}}.}$

# Calculation of tan 18°

 ${\displaystyle \tan 18^{\circ }={\frac {\sin 18^{\circ }}{\cos 18^{\circ }}}}$ ${\displaystyle ={\frac {{\sqrt {5}}-1}{4}}\cdot {\sqrt {\frac {8}{{\sqrt {5}}+5}}}}$ ${\displaystyle \tan ^{2}18^{\circ }={\frac {{\sqrt {5}}-1}{4}}\cdot {\frac {{\sqrt {5}}-1}{4}}\cdot {\frac {8}{{\sqrt {5}}+5}}}$ ${\displaystyle ={\frac {5-2{\sqrt {5}}}{5}}}$ ${\displaystyle \tan 18^{\circ }={\sqrt {\frac {5-2{\sqrt {5}}}{5}}}}$ ${\displaystyle ={\sqrt {\frac {5\cdot 5-5\cdot 2{\sqrt {5}}}{5\cdot 5}}}}$ ${\displaystyle ={\frac {\sqrt {25-10{\sqrt {5}}}}{5}}}$

# Values for 18° and 36°

x in radians x in degrees ${\displaystyle \sin(x)}$ ${\displaystyle \cos(x)}$ ${\displaystyle \tan(x)}$
${\displaystyle {\frac {\pi }{10}}}$ ${\displaystyle 18^{\circ }}$ ${\displaystyle {\frac {{\sqrt {5}}-1}{4}}}$ ${\displaystyle {\frac {\sqrt {10+2{\sqrt {5}}}}{4}}}$ ${\displaystyle {\frac {\sqrt {25-10{\sqrt {5}}}}{5}}}$
${\displaystyle {\frac {\pi }{5}}}$ ${\displaystyle 36^{\circ }}$ ${\displaystyle {\frac {\sqrt {10-2{\sqrt {5}}}}{4}}}$ ${\displaystyle {\frac {{\sqrt {5}}+1}{4}}}$ ${\displaystyle {\sqrt {5-2{\sqrt {5}}}}}$

# Calculation of cos 9°

## Using half-angle formula

 ${\displaystyle \cos ^{2}9^{\circ }={\frac {1+\cos 18^{\circ }}{2}}}$ ${\displaystyle ={\frac {1+{\sqrt {\frac {{\sqrt {5}}+5}{8}}}}{2}}}$ ${\displaystyle ={\frac {1+{\sqrt {\frac {2{\sqrt {5}}+10}{16}}}}{2}}}$ ${\displaystyle ={\frac {1+{\frac {\sqrt {2{\sqrt {5}}+10}}{4}}}{2}}}$ ${\displaystyle ={\frac {4+{\sqrt {2{\sqrt {5}}+10}}}{8}}}$ ${\displaystyle \cos 9^{\circ }={\sqrt {\frac {4+{\sqrt {2{\sqrt {5}}+10}}}{8}}}}$

## Using difference formula

π/5 36° ${\displaystyle {\frac {\sqrt {10-2{\sqrt {5}}}}{4}}}$ ${\displaystyle {\frac {1+{\sqrt {5}}}{4}}}$
π/4 45° ${\displaystyle {\frac {\sqrt {2}}{2}}}$ ${\displaystyle {\frac {\sqrt {2}}{2}}}$

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

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

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

By using similar techniques many exact values can be calculated.

# Exact Values for Common Angles

${\displaystyle 0}$ ${\displaystyle 0}$ ${\displaystyle 1}$ ${\displaystyle 0}$
π/60 ${\displaystyle {\frac {\sqrt {8-{\sqrt {3}}({\sqrt {5}}+1)-{\sqrt {10-2{\sqrt {5}}}}}}{4}}}$ ${\displaystyle {\frac {\sqrt {8+{\sqrt {3}}({\sqrt {5}}+1)+{\sqrt {10-2{\sqrt {5}}}}}}{4}}}$ ${\displaystyle {\sqrt {\frac {8-{\sqrt {3}}({\sqrt {5}}+1)-{\sqrt {10-2{\sqrt {5}}}}}{8+{\sqrt {3}}({\sqrt {5}}+1)+{\sqrt {10-2{\sqrt {5}}}}}}}}$
π/30 ${\displaystyle {\frac {{\sqrt {30-6{\sqrt {5}}}}-{\sqrt {5}}-1}{8}}}$ ${\displaystyle {\frac {{\sqrt {3}}({\sqrt {5}}+1)+{\sqrt {10-2{\sqrt {5}}}}}{8}}}$ ${\displaystyle {\frac {{\sqrt {5-2{\sqrt {5}}}}\cdot ({\sqrt {5}}+1)+{\sqrt {3}}\cdot (1-{\sqrt {5}})}{2}}}$
π/20 ${\displaystyle {\sqrt {\frac {4-{\sqrt {10+2{\sqrt {5}}}}}{8}}}}$ ${\displaystyle {\sqrt {\frac {4+{\sqrt {10+2{\sqrt {5}}}}}{8}}}}$ ${\displaystyle {\sqrt {11+4{\sqrt {5}}-(3+{\sqrt {5}}){\sqrt {10+2{\sqrt {5}}}}}}}$
π/15 12° ${\displaystyle {\frac {\sqrt {7-{\sqrt {5}}-{\sqrt {30-6{\sqrt {5}}}}}}{4}}}$ ${\displaystyle {\frac {\sqrt {9+{\sqrt {5}}+{\sqrt {30-6{\sqrt {5}}}}}}{4}}}$ ${\displaystyle {\frac {\sqrt {92-40{\sqrt {5}}+(6{\sqrt {5}}-14){\sqrt {30-6{\sqrt {5}}}}}}{2}}}$
π/12 15° ${\displaystyle {\frac {{\sqrt {6}}-{\sqrt {2}}}{4}}}$ ${\displaystyle {\frac {{\sqrt {6}}+{\sqrt {2}}}{4}}}$ ${\displaystyle 2-{\sqrt {3}}}$
π/10 18° ${\displaystyle {\frac {{\sqrt {5}}-1}{4}}}$ ${\displaystyle {\frac {\sqrt {10+2{\sqrt {5}}}}{4}}}$ ${\displaystyle {\frac {\sqrt {25-10{\sqrt {5}}}}{5}}}$
7π/60 21° ${\displaystyle {\sqrt {\frac {4-{\sqrt {7+{\sqrt {30-6{\sqrt {5}}}}-{\sqrt {5}}}}}{8}}}}$ ${\displaystyle {\sqrt {\frac {4+{\sqrt {7+{\sqrt {30-6{\sqrt {5}}}}-{\sqrt {5}}}}}{8}}}}$ ${\displaystyle {\sqrt {\frac {4-{\sqrt {7+{\sqrt {30-6{\sqrt {5}}}}-{\sqrt {5}}}}}{4+{\sqrt {7+{\sqrt {30-6{\sqrt {5}}}}-{\sqrt {5}}}}}}}}$
π/8 22.5° ${\displaystyle {\frac {\sqrt {2-{\sqrt {2}}}}{2}}}$ ${\displaystyle {\frac {\sqrt {2+{\sqrt {2}}}}{2}}}$ ${\displaystyle {\sqrt {2}}-1}$
2π/15 24° ${\displaystyle {\frac {{\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}}{8}}}$ ${\displaystyle {\frac {1+{\sqrt {5}}+{\sqrt {30-6{\sqrt {5}}}}}{8}}}$ ${\displaystyle {\frac {(3{\sqrt {5}}+7){\sqrt {5-2{\sqrt {5}}}}-({\sqrt {5}}+3){\sqrt {3}}}{2}}}$
3π/20 27° ${\displaystyle {\sqrt {\frac {4-{\sqrt {10-2{\sqrt {5}}}}}{8}}}}$ ${\displaystyle {\sqrt {\frac {4+{\sqrt {10-2{\sqrt {5}}}}}{8}}}}$ ${\displaystyle {\sqrt {11-4{\sqrt {5}}+({\sqrt {5}}-3){\sqrt {10-2{\sqrt {5}}}}}}}$
π/6 30° ${\displaystyle {\frac {1}{2}}}$ ${\displaystyle {\frac {\sqrt {3}}{2}}}$ ${\displaystyle {\frac {\sqrt {3}}{3}}}$
11π/60 33° ${\displaystyle {\frac {\sqrt {8-{\sqrt {15}}-{\sqrt {3}}+{\sqrt {10-2{\sqrt {5}}}}}}{4}}}$ ${\displaystyle {\frac {\sqrt {8+{\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}}}{4}}}$ ${\displaystyle {\sqrt {\frac {8-{\sqrt {15}}-{\sqrt {3}}+{\sqrt {10-2{\sqrt {5}}}}}{8+{\sqrt {15}}+{\sqrt {3}}-{\sqrt {10-2{\sqrt {5}}}}}}}}$
π/5 36° ${\displaystyle {\frac {\sqrt {10-2{\sqrt {5}}}}{4}}}$ ${\displaystyle {\frac {1+{\sqrt {5}}}{4}}}$ ${\displaystyle {\sqrt {5-2{\sqrt {5}}}}}$
13π/60 39° ${\displaystyle {\frac {\sqrt {8-2{\sqrt {7-{\sqrt {5}}-{\sqrt {30-6{\sqrt {5}}}}}}}}{4}}}$ ${\displaystyle {\frac {\sqrt {8+2{\sqrt {7-{\sqrt {5}}-{\sqrt {30-6{\sqrt {5}}}}}}}}{4}}}$ ${\displaystyle {\sqrt {\frac {8-2{\sqrt {7-{\sqrt {5}}-{\sqrt {30-6{\sqrt {5}}}}}}}{8+2{\sqrt {7-{\sqrt {5}}-{\sqrt {30-6{\sqrt {5}}}}}}}}}}$
7π/30 42° ${\displaystyle {\frac {\sqrt {9-{\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}}}{4}}}$ ${\displaystyle {\frac {\sqrt {7+{\sqrt {30-6{\sqrt {5}}}}-{\sqrt {5}}}}{4}}}$ ${\displaystyle {\sqrt {\frac {9-{\sqrt {30-6{\sqrt {5}}}}+{\sqrt {5}}}{7+{\sqrt {30-6{\sqrt {5}}}}-{\sqrt {5}}}}}}$
π/4 45° ${\displaystyle {\frac {\sqrt {2}}{2}}}$ ${\displaystyle {\frac {\sqrt {2}}{2}}}$ ${\displaystyle 1}$

## Other Expressions of Exact Values

 Depending on how the value is calculated, exact trigonometric values can be expressed in different ways. For example: ${\displaystyle \tan 9^{\circ }=1+{\sqrt {5}}-({\sqrt {5}}+2){\sqrt {(5-2{\sqrt {5}})}}}$ This value of ${\displaystyle \tan 9^{\circ }}$ contains calculation of 2 square roots. Value shown in table above contains calculation of 3 square roots. ${\displaystyle \tan 27^{\circ }={\sqrt {5}}-1-{\sqrt {5-2{\sqrt {5}}}}}$ ${\displaystyle \tan 33^{\circ }={\frac {(2-(2-{\sqrt {3}})(3+{\sqrt {5}}))\ \cdot \ (2+{\sqrt {2(5-{\sqrt {5}})}})}{4}}}$ This value of ${\displaystyle \tan 33^{\circ }}$ 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:

 # python code # Constants.py import decimal D = decimal.Decimal decimal.getcontext().prec = 50 L1 = [] L1 += [('Angle in Degrees', ('Sine', 'Cosine', 'Tangent'))] L1 += [(0, (0, 1, 0))] r5 = D(5).sqrt() r3 = D(3).sqrt() r2 = D(2).sqrt() root1 = (10 - 2*r5).sqrt() sval = (8 - r3*(r5+1) - root1).sqrt() cval = (8 + r3*(r5+1) + root1).sqrt() L1 += [(3, ( sval/4, cval/4, sval/cval ) )] L1 += [(6, ( ((30 - 6*r5).sqrt() - r5 - 1) / 8, ( r3*(r5 + 1) + root1 ) / 8, ( (r5+1)*(5 - 2*r5).sqrt() + r3*(1 - r5) ) / 2 ) )] v = (10 + 2*r5).sqrt() sval = 4 - v cval = 4 + v r = (5 - 2*r5).sqrt() L1 += [( 9, ( (sval/8).sqrt(), (cval/8).sqrt(), ( 11 + 4*r5 - (3+r5)*v ).sqrt() ) # (+ 1 + r5 - r*(r5 + 2)) ) )] v = (30-6*r5).sqrt() L1 += [( 12, ( (7 - r5 - v).sqrt() / 4, (9 + r5 + v).sqrt() / 4, (92 - 40*r5 + (6*r5-14)*v ).sqrt() / 2 ) )] L1 += [( 15, ( (r2*r3 - r2)/4, (r2*r3 + r2)/4, 2 - r3 ) )] L1 += [( 18, ( (r5-1)/4, ((10 + 2*r5).sqrt())/4, (25 - 10*r5).sqrt()/5 ) )] v = (7 + (30 - 6*r5).sqrt() - r5).sqrt() sval = ((4 - v)/8).sqrt() cval = ((4 + v)/8).sqrt() L1 += [( 21, ( sval, cval, sval/cval ) )] L1 += [( 22.5, ( (2 - r2).sqrt()/2, (2 + r2).sqrt()/2, r2 - 1 ) )] L1 += [( 24, ( ( r3*r5 + r3 - root1 ) / 8, (1 + r5 + (30 - 6*r5).sqrt()) / 8, ( (3*r5+7)*(5-2*r5).sqrt() - r3*(r5+3) ) / 2 ) )] v = root1 L1 += [( 27, ( ((4 - v)/8).sqrt(), ((4 + v)/8).sqrt(), ( 11 - 4*r5 + (r5-3)*v ).sqrt() ) )] L1 += [( 30, ( D('0.5'), r3/2, r3/3 ) )] sval = 8 - r3*r5 - r3 + root1 sval = sval.sqrt() cval = 8 + r3*r5 + r3 - root1 cval = cval.sqrt() L1 += [( 33, ( sval/4, cval/4, sval/cval ) )] L1 += [( 36, ( (10-2*r5).sqrt()/4, (1+r5)/4, (5 - 2*r5).sqrt() ) )] v1 = (30 - 6*r5).sqrt() v2 = 2*(7 - r5 - v1).sqrt() sval = (8 - v2).sqrt() cval = (8 + v2).sqrt() L1 += [( 39, ( sval/4, cval/4, sval/cval ) )] v = (30 - 6*r5).sqrt() sval = (9 - v + r5).sqrt()/4 cval = (7 + v - r5).sqrt()/4 L1 += [( 42, ( sval, cval, sval/cval ) )] L1 += [( 45, ( r2/2, r2/2, D(1) ) )] dict1 = dict(L1) 

### 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.

 almostZero = D('1e-' + str(decimal.getcontext().prec - 2)); print () print ('Checking each entry.') # Basic check of each entry. for angle in dict1 : if isinstance (angle, str) : continue sine, cosine, tangent = dict1[angle] print () print (angle) print (' ', sine) print (' ', cosine) print (' ', tangent) diff = abs(1 -(sine*sine + cosine*cosine)) if diff > almostZero : print ('error1:', diff) diff = abs (sine/cosine - tangent) if diff > almostZero : print ('error2:', diff) def getData (angle) : # sin,cos = getData(angle) if 135 >= angle >= 0 : pass else : print ('error3: angle =', angle) if angle > 90 : s1,c1,t1 = dict1[angle-90] return (c1, -s1) if angle > 45 : s1,c1,t1 = dict1[90-angle] return (c1,s1) s1,c1,t1 = dict1[angle] return (s1,c1) print () print ('Checking doubles and triples.') for angle1 in dict1 : if isinstance (angle1, str) : continue s1,c1,t1 = dict1[angle1] angle2 = 2*angle1 s2,c2 = getData(angle2) s2_ = 2*s1*c1 diff = abs (s2_ - s2) if diff > almostZero : print ('error4:', diff) angle3 = 3*angle1 s3,c3 = getData(angle3) c3_ = 4*c1*c1*c1 - 3*c1 diff = abs (c3_ - c3) if diff > almostZero : print ('error5:', diff) 

Testing taylor series for ${\displaystyle \arctan x}$