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

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°

Template:RoundBoxTop 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 $x=\sqrt{X}.$

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

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

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

Derivation of cos 36°

Template:RoundBoxTop 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: $\frac{1\pm \sqrt{5}}{4}.$

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

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

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

Calculation of tan 18°

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

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

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

Values for 18° and 36°

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°

Using half-angle formula

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

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

Using difference formula

Radians	Degrees	Template:Math	Template:Math
Template:Math	Template:Math	$\frac{\sqrt{10-2\sqrt{5}}}{4}$	$\frac{1+\sqrt{5}}{4}$
Template:Math	Template:Math	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$

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

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

Exact Values for Common Angles

Radians	Degrees	Template:Math	Template:Math	Template:Math
$0$	Template:Math	$0$	$1$	$0$
Template:Math	Template:Math	$\frac{\sqrt{8-\sqrt{3}(\sqrt{5}+1)-\sqrt{10-2\sqrt{5}}}}{4}$	$\frac{\sqrt{8+\sqrt{3}(\sqrt{5}+1)+\sqrt{10-2\sqrt{5}}}}{4}$	$\sqrt{\frac{8-\sqrt{3}(\sqrt{5}+1)-\sqrt{10-2\sqrt{5}}}{8+\sqrt{3}(\sqrt{5}+1)+\sqrt{10-2\sqrt{5}}}}$
Template:Math	Template:Math	$\frac{\sqrt{30-6\sqrt{5}}-\sqrt{5}-1}{8}$	$\frac{\sqrt{3}(\sqrt{5}+1)+\sqrt{10-2\sqrt{5}}}{8}$	$\frac{\sqrt{5-2\sqrt{5}}\cdot (\sqrt{5}+1)+\sqrt{3}\cdot (1-\sqrt{5})}{2}$
Template:Math	Template:Math	$\sqrt{\frac{4-\sqrt{10+2\sqrt{5}}}{8}}$	$\sqrt{\frac{4+\sqrt{10+2\sqrt{5}}}{8}}$	$\sqrt{11+4\sqrt{5}-(3+\sqrt{5})\sqrt{10+2\sqrt{5}}}$
Template:Math	Template:Math	$\frac{\sqrt{7-\sqrt{5}-\sqrt{30-6\sqrt{5}}}}{4}$	$\frac{\sqrt{9+\sqrt{5}+\sqrt{30-6\sqrt{5}}}}{4}$	$\frac{\sqrt{92-40\sqrt{5}+(6\sqrt{5}-14)\sqrt{30-6\sqrt{5}}}}{2}$
Template:Math	Template:Math	$\frac{\sqrt{6}-\sqrt{2}}{4}$	$\frac{\sqrt{6}+\sqrt{2}}{4}$	$2-\sqrt{3}$
Template:Math	Template:Math	$\frac{\sqrt{5}-1}{4}$	$\frac{\sqrt{10+2\sqrt{5}}}{4}$	$\frac{\sqrt{25-10\sqrt{5}}}{5}$
Template:Math	Template:Math	$\sqrt{\frac{4-\sqrt{7+\sqrt{30-6\sqrt{5}}-\sqrt{5}}}{8}}$	$\sqrt{\frac{4+\sqrt{7+\sqrt{30-6\sqrt{5}}-\sqrt{5}}}{8}}$	$\sqrt{\frac{4-\sqrt{7+\sqrt{30-6\sqrt{5}}-\sqrt{5}}}{4+\sqrt{7+\sqrt{30-6\sqrt{5}}-\sqrt{5}}}}$
Template:Math	Template:Math	$\frac{\sqrt{2-\sqrt{2}}}{2}$	$\frac{\sqrt{2+\sqrt{2}}}{2}$	$\sqrt{2}-1$
Template:Math	Template:Math	$\frac{\sqrt{15}+\sqrt{3}-\sqrt{10-2\sqrt{5}}}{8}$	$\frac{1+\sqrt{5}+\sqrt{30-6\sqrt{5}}}{8}$	$\frac{(3\sqrt{5}+7)\sqrt{5-2\sqrt{5}}-(\sqrt{5}+3)\sqrt{3}}{2}$
Template:Math	Template:Math	$\sqrt{\frac{4-\sqrt{10-2\sqrt{5}}}{8}}$	$\sqrt{\frac{4+\sqrt{10-2\sqrt{5}}}{8}}$	$\sqrt{11-4\sqrt{5}+(\sqrt{5}-3)\sqrt{10-2\sqrt{5}}}$
Template:Math	Template:Math	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{3}}{3}$
Template:Math	Template:Math	$\frac{\sqrt{8-\sqrt{15}-\sqrt{3}+\sqrt{10-2\sqrt{5}}}}{4}$	$\frac{\sqrt{8+\sqrt{15}+\sqrt{3}-\sqrt{10-2\sqrt{5}}}}{4}$	$\sqrt{\frac{8-\sqrt{15}-\sqrt{3}+\sqrt{10-2\sqrt{5}}}{8+\sqrt{15}+\sqrt{3}-\sqrt{10-2\sqrt{5}}}}$
Template:Math	Template:Math	$\frac{\sqrt{10-2\sqrt{5}}}{4}$	$\frac{1+\sqrt{5}}{4}$	$\sqrt{5-2\sqrt{5}}$
Template:Math	Template:Math	$\frac{\sqrt{8-2\sqrt{7-\sqrt{5}-\sqrt{30-6\sqrt{5}}}}}{4}$	$\frac{\sqrt{8+2\sqrt{7-\sqrt{5}-\sqrt{30-6\sqrt{5}}}}}{4}$	$\sqrt{\frac{8-2\sqrt{7-\sqrt{5}-\sqrt{30-6\sqrt{5}}}}{8+2\sqrt{7-\sqrt{5}-\sqrt{30-6\sqrt{5}}}}}$
Template:Math	Template:Math	$\frac{\sqrt{9-\sqrt{30-6\sqrt{5}}+\sqrt{5}}}{4}$	$\frac{\sqrt{7+\sqrt{30-6\sqrt{5}}-\sqrt{5}}}{4}$	$\sqrt{\frac{9-\sqrt{30-6\sqrt{5}}+\sqrt{5}}{7+\sqrt{30-6\sqrt{5}}-\sqrt{5}}}$
Template:Math	Template:Math	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	$1$

Other Expressions of Exact Values

Template:RoundBoxTop 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. Template:RoundBoxBottom

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: Template:RoundBoxTop

# 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. Template:RoundBoxTop

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)

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 $\arctan x$