Sines and Cosines for Fractions of π
The exact values of the trig ratios sin x and cos x for x = π, π/2, π/3, π/4 and π/6 are
taught in schools i.e.
x
π π/2 π/3 π/4 π/6
sin x
0
1
cos x
−1
0
√
3
2
1
2
√1
2
√1
2
1
√2
3
2
But what about sin π5 and cos π5 ? These trig ratios can be obtained by finding expressions
for sin 5x and cos 5x in terms of powers of sin x and cos x i.e.
cos 5x = 5 cos x − 20 cos3 x + 16 cos5 x
sin 5x = 5 sin x − 20 sin3 x + 16 sin5 x
. Setting sin 5x = 0 gives a fifth-order equation for sin x but when a factor of sin x is canceled,
this gives a quadratic equation for sin2 x and hence leads
to the values for sin π5 and
sin 2π
. A
5
q
q
√
√
similar equation can be solved to get cos π5 . sin π5 = 41 10 − 2 5 and cos π5 = 14 6 + 2 5 =
√
1
1
5. The values of sin 2π
+
and cos 2π
can also be found from the analysis.
4
4
5
5
Exact values for sin π8 and cos π8 can be found by solving the equations 1 − 2 sin2 x = √12
q
³ ´
√
and 2 cos2 x − 1 = √12 (both based on cos 2x = cos π4 . These give sin π8 = 21 2 − 2 and
q
√
π
cos π8 = 12 2 + 2. This is an example of finding the trig ratios for 2n
if the values for πn are
known
and cos 2π
Using this principle, the values of sin and cos of π/10 can be obtained from sin 2π
5
5
as found earlier.
Values of sin and cos of π/12 can be obtained by two methods i.e. one involving the sin
and cos of π/3 and π/4 and the other involving the ratios at π/6. The first of these states that
sin π/12 = sin π/3 cos π/4 + sin π/3 cos π/4.
So, it would seem that there are two circumstances under which sin π/n and cos π/n can be
found.
1). If n = 2m i.e. if n is even and sin π/m and cos π/m are known then sin π/n and cos π/n
can be found e.g. sin and cos of π/10 are known so sin and cos of π/20 can be found.
2). If 1/n can be expressed as 1/a − 1/b where the trig ratios can be calculated for π/a and
π/b, then the ratios can be calculated for π/n. For example, as the trig ratios can be calculated
for π/4 and π/5 so they can be calculated for π/4 − π/5.
1
However, in practice, route 2) will only give a few values that cannot be found through route
1). There is however, another circumstance under which sin and cos of π/n can be found. If
k
n is a Fermat Prime i.e. a prime number of the form 22 + 1, then sin π/n and cos π/n can be
found exactly. n = 3 and n = 5 are Fermat primes and the sine and cosine of π/3 and π/5 can
be found (see above).
The next Fermat prime is 17 and it turns out that sin π/17 and cos π/17 can be found. Similarly, for the two remaining known Fermat primes n = 257 and n = 65537, the sine and cosine
of π/n can be found.
Using technique 1) above, on the basis of sinπ/17 and cos π/17, the sines and cosines of π/34,
π/68, π/136 etc. can be found.
Route 2) can be used for sin π/n and cos π/n where n is the product of two Fermat Primes
greater than 2.
Also, it is worth pointing out that if sin π/n and cos π/n can be found, then so can sin mπ/n
and cos mπ/n from the formulae for sin and cos of A + B etc.
So, sin π/n and cos π/n can be found for
n = 2k × 3p × 5q × 17r × 257s × 65537t where k is a positive or zero integer and the indices
p, q, r, s, t are all either zero or 1. Values of n under 1000 are as given in the following table.
2
3
4
5
6
8
10 12
15
16
17
20
24
30
32
34
40
48
51
60
64
68
80
85
96
102
120
128
136
160
170
192
204
240
255
256
257
272
320
340
384
408
480
510
512
514
544
640
680 768 771 816 960
The sines and cosines of values of mπ/n for n ≤ 32 are given in the following tables.
2
sin π = 0
cos π = −1
sin π2 = 1
3
2
π
√1
4 = q
2
√
π
1
=
10
−
2
5
5
4 q
√
1
2π
5 = 4 10 + 2 5
π
1
6 = 2q
√
1
π
2
−
=
2
8
2 q
√
3π
1
2
8 = 2q 2 +
√
√
1
1
π
1
6
−
2
=
+
5
=
−
10
4q
4
4 5
√
√
3π
1
1
1
10 = 4 6 + 2 5 = 4 + 4 5
³√
√ ´
1
π
6
−
2
=
12
4
´
³√
√
1
5π
6
+
2
=
12
4q
√
√
√
1
1
π
1
10
+
2
3
−
=
5
+
15
8 q
8
8 15
√
√
√
1
1
2π
1
10
−
2
3
+
=
−
5
+
15
8
8 15
q8
√
√
√
4π
1
1
1
15 = 8 q10 + 2 5 − 8 3 + 8 15
√
√
√
7π
1
1
1
10
−
2
=
5
+
3
+
15
8r
8
8 15
q
√
π
1
=
2
−
2
+
2
16
2r
q
√
1
3π
2
16 = 2 r2 − 2 −
q
√
1
5π
2
+
2
−
=
2
16
2r
q
√
1
7π
2
16 = 2 2 + 2 +
sin π3 =
sin
sin
sin
sin
sin
sin
sin
sin
sin
sin
sin
sin
sin
sin
sin
sin
sin
sin
cos π2 = 0
√
3
1
2
π
√1
4 = q
2
√
√
π
1
1
1
=
6
+
2
5
=
+
5
4 q
4
4 5
√
√
1
1
2π
1
=
+
6
−
2
5
=
−
5
4
4 5
√4
π
3
3 = 2q
√
1
π
2
+
2
=
8
2 q
√
3π
1
2
8 = 2q 2 −
√
π
1
10
+
2
=
5
10
4q
√
3π
1
10 = 4 10 − 2 5
³√
√ ´
1
π
6
+
2
=
12
4
´
³√
√
1
5π
6
−
2
=
12
4
√ q
√
√
π
1
1
3
10
+
2
=
−
+
5
+
5
15
8
8
√ 8q
√
√
2π
1
1
3
15 = 8 + 8 5 + √8 q10 − 2 5
√
√
3
4π
1
1
10
+
2
5
+
5
=
−
15
8
8
8√ q
√
√
7π
1
1
3
10 − 2 5
15 = −r8 − 8 5 + 8
q
√
π
1
=
2
+
2
+
2
16
2r
q
√
1
3π
2
16 = 2 r2 + 2 −
q
√
1
5π
2
−
2
−
2
=
16
2r
q
√
1
7π
2
16 = 2 2 − 2 +
cos π3 =
cos
cos
cos
cos
cos
cos
cos
cos
cos
cos
cos
cos
cos
cos
cos
cos
cos
cos
q
ǫ = 17 +
√
17
√
ǫ∗ = 17 − 17
√
δ = 17 − 1
q
√
√
√
α = 34 + 6 × 17 + 2 × δ × ǫ∗ − 8 × 2 × ǫ
q
√
√
√
β = 2 17 + 3 17 − 2 2ǫ − 2ǫ∗
√
√ q
= 18 2 (ǫ∗ )2 − 2(α + ǫ∗ )
√
√ q
√
π
= 81 2 15 + 17 + 2(α + ǫ∗ )
cos 17
√
√ q
√
√
1
= 16 2 4(ǫ∗ )2 − 2 2δǫ∗ + 8 2ǫ − ( 2δ + 2ǫ∗ )α
√
1
=
[δ
+
cos 2π
2(α + ǫ∗ )]
17
16
q
π
sin 17
sin 2π
17
2π
π
2π
π
sin 3π
17 = sin 17 cos 17 + cos 17 sin 17
sin 4π
17
π
π
cos 3π
= cos 2π
cos 17
− sin 2π
sin 17
17
17
17
q
√
√
√
√
1
= 128
[ 2δ + 2(α + ǫ∗ )] 4(ǫ∗ )2 − 2 2δǫ∗ + 8 2ǫ − ( 2δ + 2ǫ∗ )α
i
h
√
1
∗
∗ 2
2
2δ(α
+
ǫ
)
+
2(α
+
ǫ
)
−
128
cos 4π
=
δ
+
2
17
128
4π
π
4π
π
sin 5π
17 = sin 17 cos 17 + cos 17 sin 17
4π
π
4π
π
cos 5π
17 = cos 17 cos 17 − sin 17 sin 17
4π
2π
4π
2π
sin 6π
17 = sin 17 cos 17 + cos 17 sin 17
4π
2π
4π
2π
cos 6π
17 = cos 17 cos 17 − sin 17 sin 17
8π
π
8π
π
sin 7π
17 = sin 17 cos 17 − cos 17 sin 17
sin 8π
17
π
π
cos 7π
= cos 8π
cos 17
+ sin 8π
sin 17
17
17
17
q
√
√
√
√
√
1
= 16 136 − 8 17 + 8 2ǫ − 2( 34 − 3 2)ǫ∗ + 2β(δ + 2ǫ∗ )
q
√ ∗
√
√
√
8π
1
cos 17 = 16 [δ + 2ǫ − 2 17 + 3 17 − 2ǫ∗ − 2 2ǫ]
4
√
sin =
− 2 10 + 2 5
q
√
1
3π
sin 20 = 4 8 − 2 10 − 2 5
r
q
√
7π
1
sin 20 = 4 8 + 2 10 − 2 5
r
q
√
9π
1
sin 20 = 4 8 + 2 10 + 2 5
q
√
√
1
π
sin 24 = 4 8 − 2 2 − 2 6
q
√
√
5π
1
sin 24 = 4 8 + 2 2 − 2 6
q
√
√
7π
1
sin 24 = 4 8 − 2 2 + 2 6
q
√
√
1
=
sin 11π
8
+
2
2
+
2
6
24
4r
q
√
√
π
= 18 −1 − 5 + 30 − 6 5
sin 30
r
q
√
√
7π
1
sin 30 = 8 1 − 5 + 30 + 6 5
r
q
√
√
11π
1
sin 30 = 8 1 + 5 + 30 − 6 5
r
q
√
√
13π
1
sin 30 = 8 −1 + 5 + 30 + 6 5
s
r
q
√
1
π
sin 32 = 2 2 − 2 + 2 + 2
s
r
q
√
3π
1
sin 32 = 2 2 − 2 + 2 − 2
s
r
q
√
5π
1
sin 32 = 2 2 − 2 − 2 − 2
s
r
q
√
7π
1
sin 32 = 2 2 − 2 − 2 + 2
s
r
q
√
1
9π
sin 32 = 2 2 + 2 − 2 + 2
s
r
q
√
1
11π
sin 32 = 2 2 + 2 − 2 − 2
s
r
q
√
13π
1
sin 32 = 2 2 + 2 + 2 − 2
s
r
q
√
15π
1
sin 32 = 2 2 + 2 + 2 + 2
π
20
r
1
4 r8
q
√
+ 2 10 + 2 5
q
√
1
3π
cos 20 = 4 8 + 2 10 − 2 5
r
q
√
7π
1
cos 20 = 4 8 − 2 10 − 2 5
r
q
√
9π
1
cos 20 = 4 8 − 2 10 + 2 5
q
√
√
1
π
cos 24 = 4 8 + 2 2 + 2 6
q
√
√
5π
1
cos 24 = 4 8 − 2 2 + 2 6
q
√
√
7π
1
cos 24 = 4 8 + 2 2 − 2 6
q
√
√
1
=
cos 11π
8
−
2
2
−
2
6
24
4r
q
√
√
√
π
= 81
cos 30
3 + 15 + 10 − 2 5
r
q
√
√
√
7π
1
cos 30 = 8 − 3 + 15 + 10 + 2 5
r
q
√
√
√
11π
1
3 + 15 − 10 − 2 5
cos 30 = 8
r
q
√
√
√
13π
1
3 − 15 + 10 + 2 5
cos 30 = 8
s
r
q
√
1
π
cos 32 = 2 2 + 2 + 2 + 2
s
r
q
√
3π
1
cos 32 = 2 2 + 2 + 2 − 2
s
r
q
√
5π
1
cos 32 = 2 2 + 2 − 2 − 2
s
r
q
√
7π
1
cos 32 = 2 2 + 2 − 2 + 2
s
r
q
√
1
9π
cos 32 = 2 2 − 2 − 2 + 2
s
r
q
√
1
11π
cos 32 = 2 2 − 2 − 2 − 2
s
r
q
√
13π
1
cos 32 = 2 2 − 2 + 2 − 2
s
r
q
√
15π
1
cos 32 = 2 2 − 2 + 2 + 2
cos
5
π
20
=
r
1
4 r8
q