# document.write (document.title)

 Math Help > Trigonometry > Trig Equivalences > Special Angles

# Special Angles

Special angles, or common angles, which have sines and cosines that are roots of polynomials.

 Consider this figure. Circle ABC has center D.  AD, BD, and CD are radii.  AC is a diameter.  BD ^ AC. DE = EC, and that length is denoted a. EF = BE, and that length is denoted x+a.

We will set out to prove the following:

CD = 2a is the side of an inscribed regular hexagon.
DF = x is the side of an inscribed regular decagon.
BF = y is the side of an inscribed regular pentagon.

 CF x DF + ED² = EF² because (2a+x)x + a² = (a+x)² CF x DF + ED² = BE² " EF = BE DB² + ED² = BE² " Pythagorean Theorem CF x DF = DB² " DB² in the eq above takes the place of CF x DF in the eq above that CF x DF = DC² " DB and DC are both radii, and equal DF:DC is the golden ratio sqrt(5)-1 : 2 " Look at triangle BDE. r=2a, so the sides are a and 2a, and the hypotenuse is x+a. (x+a)² = a²+(2a)² = 5a² x+a = sqrt(5)a,  so  x = (sqrt(5)-1)a, and since  r=2a,  we get... x:r = (sqrt(5)-1) : 2 \DC = r is the length of the side of an inscribed hexagon " DC is a radius; r is the side of a hexagon \DF = x is the length of the side of an inscribed decagon " Euclid, "the side of the hexagon and the side of a decagon which are inscribed in the same circle... cut that line in the extreme and mean ratio (Euclid XIII, 9)" (Ptolemy 20) a.k.a. the golden ratio.  i.e. the ratio of x:r \BF = y is the length of the side of an inscribed pentagon " Euclid again, "the square on the side of a pentagon is equal to the square on the side of the hexagon together with the square on the side of a decagon, all inscribed on the same circle. (Euclid XIII, 10)" (Ptolemy 20) i.e. x²+r²=y² sin(36�) = sqrt(10-2sqrt(5))/4 " Let r=2.  Then y is the side of an inscribed pentagon. a=1, so x=sqrt(5)-1, so x²=6-2sqrt(5) y=sqrt(x²+r²)=sqrt(10-2sqrt(5)) Half the side of the pentagon divided by the radius is the sin of half the central angle, so sin 36� = y/4 = sqrt(10-2sqrt(5))/4

Here is a table of "exact" values of common values of sin and cos.  I started with the standard values associated with the 30� 60� 90� triangle and the 45� 45� 90� triangle, and used the sum-of-sin, double angle and half angle identities to derive sines of other angles.  Then I folded in Ptolemy's contribution, which is sin 36�.

In many of these expressions, we find sqrt(c+sqrt(d)) for two rational numbers, c and d.  Often c+sqrt(d) is the perfect square of an expression of the form sqrt(a)+sqrt(b).  Finding these is tricky.  See Simplifying Nested Radicals for more info on this.

 sin(30�) = cos(60�) = sqrt(1/4) = 1/2 sin(45�) = cos(45�) = sqrt(1/2) sin(60�) = cos(30�) = sqrt(3/4) sin(15�) = cos(75�) = sin 45� cos 30� - cos 45� sin 30� = sqrt(3/8)-sqrt(1/8) sin(75�) = cos(15�) = sin 45� cos 30� + cos 45� sin 30� = sqrt(3/8)+sqrt(1/8) sin(36�) = cos(54�) = sqrt(10-2sqrt(5))/4 = sqrt(5/8-sqrt(5)/8) sin(54�) = cos(36�) = sqrt(1-sin² 36�) = sqrt(3/8+sqrt(5)/8) = sqrt(5/16)+1/4 sin(6�) = cos(84�) = sin 36� cos 30� - cos 36� sin 30� = sqrt(15/32-sqrt(45)/32)-sqrt(5/64)-1/8 sin(84�) = cos(6�) = sin 54� cos 30� + cos 54� sin 30� = sqrt(15)/8+sqrt(3)/8+sqrt(5/32-sqrt(5)/32) sin(18�) = cos(72�) = sin 54� cos 36� - cos 54� sin 36� = sin² 54� - sin² 36� = (3/8+sqrt(5)/8)-(5/8-sqrt(5)/8) = sqrt(5/16)-1/4 Note: this is exactly half the Golden Ratio sin(72�) = cos(18�) = sqrt(1-sin² 18�) = sqrt(1-5/16+sqrt(5)/8-1/16) = sqrt(5/8+sqrt(5)/8) sin(3�) = cos(87�) = sqrt((1-cos 6�)/2) = sqrt((1-(sqrt(9/32+sqrt(45)/32)+sqrt(5/32-sqrt(5)/32)))/2) = sqrt(1/2-sqrt(9/128+sqrt(45)/128)-sqrt(5/128-sqrt(5)/128)) = sqrt(1/2-sqrt(3)/16-sqrt(15)/16-sqrt(5/128-sqrt(5)/128)) or (this one has only two levels of sqrt nesting)... sin 75� cos 72� - cos 75� sin 72� = (sqrt(3/8)+sqrt(1/8))*(sqrt(5/16)-1/4)-(sqrt(3/8)-sqrt(1/8))*(sqrt(5/8+sqrt(5)/8)) = sqrt(15/128)+sqrt(5/128)-sqrt(3/128)-sqrt(1/128)-sqrt(3/8)*(sqrt(5/8+sqrt(5)/8))+sqrt(1/8)*(sqrt(5/8+sqrt(5)/8)) = sqrt(15/128)+sqrt(5/128)-sqrt(3/128)-sqrt(1/128)-sqrt(15/64+sqrt(45)/64)+sqrt(5/64+sqrt(5)/64) sin(87�) = cos(3�) = sqrt((1+cos 6�)/2) = sqrt(1/2+sqrt(3)/16+sqrt(15)/16+sqrt(5/128-sqrt(5)/128)) sin(27�) = cos(63�) = sqrt((1-cos 54�)/2) = sqrt(1/2-sqrt(5/32-sqrt(5)/32)) = sqrt(5/16+sqrt(5)/16)-sqrt(3/16-sqrt(5)/16) = sqrt(5/16+sqrt(5)/16)-sqrt(5/32)+sqrt(1/32) sin(63�) = cos(27�) = sqrt((1+cos 54�)/2) = sqrt(1/2+sqrt(5/32-sqrt(5)/32)) = sqrt(5/16+sqrt(5)/16)+sqrt(3/16-sqrt(5)/16) = sqrt(5/16+sqrt(5)/16)+sqrt(5/32)-sqrt(1/32) sin(9�) = cos(81�) = sqrt((1-cos 18�)/2) = sqrt(1/2-sqrt(5/32+sqrt(5)/32)) = sqrt(3/16+sqrt(5)/16)-sqrt(5/16-sqrt(5)/16) = sqrt(1/32)+sqrt(5/32)-sqrt(5/16-sqrt(5)/16) sin(81�) = cos(9�) = sqrt((1+cos 18�)/2) = sqrt(1/2+sqrt(5/32+sqrt(5)/32)) = sqrt(3/16+sqrt(5)/16)+sqrt(5/16-sqrt(5)/16) = sqrt(1/32)+sqrt(5/32)+sqrt(5/16-sqrt(5)/16)

In Trig functions of special angles, part 2 I had hoped to find an arithmetic expression that gives the cosine of 40�, because it's one of the roots of 8x3-6x+1, but, alas, it seems that such an expression eludes me.

### Summary of exact values of sin, cos of multiples of 3 degrees

 sin 0 = cos 90 = 0 sin 3 = cos 87 = (1/16)*(-sqrt(2)-sqrt(6)+sqrt(10)+sqrt(30)+(2-2*sqrt(3))*(sqrt(5+sqrt(5)))) sin 6 = cos 84 = (1/8)*(-1-sqrt(5)+sqrt(30-6*sqrt(5))) sin 9 = cos 81 = (1/8)*(sqrt(2)+sqrt(10)-2*(sqrt(5-sqrt(5)))) sin 12 = cos 78 = (1/8)*(sqrt(3)-sqrt(15)+sqrt(10+2*sqrt(5))) sin 15 = cos 75 = (1/4)*(sqrt(6)-sqrt(2)) sin 18 = cos 72 = (1/4)*(-1+sqrt(5)) sin 21 = cos 69 = (1/16)*(sqrt(2)-sqrt(6)+sqrt(10)-sqrt(30)+(2+2*sqrt(3))*(sqrt(5-sqrt(5)))) sin 24 = cos 66 = (1/8)*(sqrt(3)+sqrt(15)-sqrt(10-2*sqrt(5))) sin 27 = cos 63 = (1/8)*(sqrt(2)-sqrt(10)+2*sqrt(5+sqrt(5))) sin 30 = cos 60 = 1/2 sin 33 = cos 57 = (1/16)*(-sqrt(2)-sqrt(6)+sqrt(10)+sqrt(30)-(2-2*sqrt(3))*(sqrt(5+sqrt(5)))) sin 36 = cos 54 = (1/4)*(sqrt(10-2*sqrt(5))) sin 39 = cos 51 = (1/16)*(sqrt(2)+sqrt(6)+sqrt(10)+sqrt(30)+(2-2*sqrt(3))*(sqrt(5-sqrt(5)))) sin 42 = cos 48 = (1/8)*(1-sqrt(5)+sqrt(30+6*sqrt(5))) sin 45 = cos 45 = (1/2)*sqrt(2) sin 48 = cos 42 = (1/8)*(-sqrt(3)+sqrt(15)+sqrt(10+2*sqrt(5))) sin 51 = cos 39 = (1/16)*(-sqrt(2)+sqrt(6)-sqrt(10)+sqrt(30)+(2+2*sqrt(3))*(sqrt(5-sqrt(5)))) sin 54 = cos 36 = (1/4)*(1+sqrt(5)) sin 57 = cos 33 = (1/16)*(-sqrt(2)+sqrt(6)+sqrt(10)-sqrt(30)+(2+2*sqrt(3))*(sqrt(5+sqrt(5)))) sin 60 = cos 30 = (1/2)*sqrt(3) sin 63 = cos 27 = (1/8)*(-sqrt(2)+sqrt(10)+2*sqrt(5+sqrt(5))) sin 66 = cos 24 = (1/8)*(1+sqrt(5)+sqrt(30-6*sqrt(5))) sin 69 = cos 21 = (1/16)*(sqrt(2)+sqrt(6)+sqrt(10)+sqrt(30)-(2-2*sqrt(3))*(sqrt(5-sqrt(5)))) sin 72 = cos 18 = (1/4)*(sqrt(10+2*sqrt(5))) sin 75 = cos 15 = (1/4)*(sqrt(6)+sqrt(2)) sin 78 = cos 12 = (1/8)*(-1+sqrt(5)+sqrt(30+6*sqrt(5))) sin 81 = cos 9 = (1/8)*(sqrt(2)+sqrt(10)+2*(sqrt(5-sqrt(5)))) sin 84 = cos 6 = (1/8)*(sqrt(3)+sqrt(15)+sqrt(10-2*sqrt(5))) sin 87 = cos 3 = (1/16)*(sqrt(2)-sqrt(6)-sqrt(10)+sqrt(30)+(2+2*sqrt(3))*(sqrt(5+sqrt(5)))) sin 90 = cos 0 = 1

### Internet references

http://hypertextbook.com/eworld/chords.shtml, which cites Ptolemy's On the Size of Chords Inscribed in a Circle (2nd Century AD).

### Related Pages in this website

Common Angles -- a way for trig students to remember the sines and cosines of the most common angles.

Trig Equivalences

Trig functions of special angles, part 2 -- cos 40� is one of the roots of 8x3-6x+1, so can we find an arithmetic expression for cos 40�?

Sin or Cos 3x, 4x, etc. -- trig functions of any multiple of an angle.

Golden Ratio -- (sqrt(5)+1)/2, a special number that comes up in a variety of geometrical contexts

The webmaster and author of this Math Help site is Graeme McRae.