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

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

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

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

From the right isosceles triangle:

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

From the equilateral triangle:

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

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

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

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

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

Derivation of cos 18°[edit | edit source]

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°[edit | edit source]

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°[edit | edit source]

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

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

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

Values for 18° and 36°[edit | edit source]

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

Calculation of cos 9°[edit | edit source]

Using half-angle formula[edit | edit source]

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

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

Using difference formula[edit | edit source]

Radians	Degrees	$sin$	$cos$
$.mw-parser-output .sfrac{white-space:nowrap}.mw-parser-output .sfrac.tion,.mw-parser-output .sfrac .tion{display:inline-block;vertical-align:-0.5em;font-size:85%;text-align:center}.mw-parser-output .sfrac .num,.mw-parser-output .sfrac .den{display:block;line-height:1em;margin:0 0.1em}.mw-parser-output .sfrac .den{border-top:1px solid}.mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px}π/5$	$36°$	${\frac {\sqrt {10-2{\sqrt {5}}}}{4}}$	${\frac {1+{\sqrt {5}}}{4}}$
$π / 4$	$45°$	${\frac {\sqrt {2}}{2}}$	${\frac {\sqrt {2}}{2}}$

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

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

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[edit | edit source]

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

Other Expressions of Exact Values[edit | edit source]

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[edit | edit source]

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[edit | edit source]

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)