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### Definitions of Trigonometric Functions

Draw a unit circle with center O. Let a central angle with initial side OP and terminal side OQ contain x radians (that is, the arc PQ has length x). Drop a perpendicular from Q to OP meeting it at R. Then OR = cos(x) and RQ = sin(x). If those directed line segments are up or to the right, the lengths are positive. If they are down or to the left, the lengths are negative.

Values at special angles:

```  x      sin(x)     cos(x)      tan(x)      cot(x)       sec(x)       csc(x)

0        0          1           0          ---           1           ---
/6      1/2      sqrt(3)/2   sqrt(3)/3   sqrt(3)     2 sqrt(3)/3      2
/4   sqrt(2)/2   sqrt(2)/2      1           1         sqrt(2)      sqrt(2)
/3   sqrt(3)/2     1/2       sqrt(3)     sqrt(3)/3       2       2 sqrt(3)/3
/2       1          0          ---          0           ---           1
2/3   sqrt(3)/2    -1/2      -sqrt(3)    -sqrt(3)/3      -2       2 sqrt(3)/3
3/4   sqrt(2)/2  -sqrt(2)/2     -1          -1        -sqrt(2)      sqrt(2)
5/6      1/2     -sqrt(3)/2  -sqrt(3)/3  -sqrt(3)    -2 sqrt(3)/3      2
0         -1           0          ---          -1           ---

```
More values at special angles:
```  x              /10                  /5
sin(x)      (-1+sqrt[5])/4      sqrt(10-2 sqrt[5])/4
cos(x)   sqrt(10+2 sqrt[5])/4      (1+sqrt[5])/4
tan(x)    sqrt(1-2/sqrt[5])      sqrt(5-2 sqrt[5])
cot(x)    sqrt(5+2 sqrt[5])      sqrt(1+2/sqrt[5])
sec(x)    sqrt(2-2/sqrt[5])         -1+sqrt[5]
csc(x)        1+sqrt[5]          sqrt(2+2/sqrt[5])

```
Use the above values and the identities below to obtain values of trigonometric functions of the following multiples of /10:

```3/10 = /2 - /5,
2/5  = /2 - /10,
3/5  = /2 + /10,
7/10 = /2 + /5,
4/5  =    - /5,
9/10 =    - /10.
```

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### Bounds

```   |sin(x)| <= 1,
|cos(x)| <= 1,
|sec(x)| >= 1,
|csc(x)| >= 1.
```

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### Identities

```   sec(x) = 1/cos(x),
csc(x) = 1/sin(x),
cot(x) = 1/tan(x),
tan(x) = sin(x)/cos(x),
cot(x) = cos(x)/sin(x).

sin(-x) = -sin(x),
cos(-x) = cos(x),
tan(-x) = -tan(x),
cot(-x) = -cot(x),
sec(-x) = sec(x),
csc(-x) = -csc(x).

sin(/2-x) = cos(x),
cos(/2-x) = sin(x),
tan(/2-x) = cot(x),
cot(/2-x) = tan(x),
sec(/2-x) = csc(x),
csc(/2-x) = sec(x).

sin(/2+x) = cos(x),
cos(/2+x) = -sin(x),
tan(/2+x) = -cot(x),
cot(/2+x) = -tan(x),
sec(/2+x) = -csc(x),
csc(/2+x) = sec(x).

sin(-x) = sin(x),
cos(-x) = -cos(x),
tan(-x) = -tan(x),
cot(-x) = -cot(x),
sec(-x) = -sec(x),
csc(-x) = csc(x).

sin(+x) = -sin(x),
cos(+x) = -cos(x),
tan(+x) = tan(x),
cot(+x) = cot(x),
sec(+x) = -sec(x),
csc(+x) = -csc(x).

sin(2+x) = sin(x),
cos(2+x) = cos(x),
tan(2+x) = tan(x),
cot(2+x) = cot(x),
sec(2+x) = sec(x),
csc(2+x) = csc(x).

sin2(x) + cos2(x) = 1,
tan2(x) + 1 = sec2(x),
1 + cot2(x) = csc2(x).

sin(x+y) = sin(x)cos(y) + cos(x)sin(y),
cos(x+y) = cos(x)cos(y) - sin(x)sin(y),
tan(x+y) = [tan(x)+tan(y)]/[1-tan(x)tan(y)],
cot(x+y) = [cot(x)cot(y)-1]/[cot(x)+cot(y)].

sin(x-y) = sin(x)cos(y) - cos(x)sin(y),
cos(x-y) = cos(x)cos(y) + sin(x)sin(y),
tan(x-y) = [tan(x)-tan(y)]/[1+tan(x)tan(y)],
cot(x-y) = [cot(x)cot(y)+1]/[cot(y)-cot(x)].

sin(2x) = 2 sin(x)cos(x),
cos(2x) = cos2(x) - sin2(x),
= 2 cos2(x) - 1,
= 1 - 2 sin2(x),
tan(2x) = [2 tan(x)]/[1-tan2(x)],
cot(2x) = [cot2(x)-1]/[2 cot(x)].

|sin(x/2)| = sqrt([1-cos(x)]/2),

|cos(x/2)| = sqrt([1+cos(x)]/2),

|tan(x/2)| = sqrt([1-cos(x)]/[1+cos(x)]),

tan(x/2) = [1-cos(x)]/sin(x),
= sin(x)/[1+cos(x)].

sin(3x) = 3 sin(x) - 4 sin3(x),
cos(3x) = 4 cos3(x) - 3 cos(x),
tan(3x) = [3 tan(x)-tan3(x)]/[1-3 tan2(x)].

sin(4x) = 4 sin(x)cos(x)[2 cos2(x)-1],
cos(4x) = 8 cos4(x) - 8 cos2(x) + 1.

sin(5x) = 5 sin(x) - 20 sin3(x) + 16 sin5(x),
cos(5x) = 16 cos5(x) - 20 cos3(x) + 5 cos(x).

sin(6x) = 2 sin(x)cos(x)[16 cos4(x) - 16 cos2(x) + 3],
cos(6x) = 32 cos6(x) - 48 cos4(x) + 18 cos2(x) - 1.

sin(nx) = 2 sin([n-1]x)cos(x) - sin([n-2]x),
cos(nx) = 2 cos([n-1]x)cos(x) - cos([n-2]x),
tan(nx) = (tan[(n-1)x]+tan[x])/(1-tan[(n-1)x]tan[x]).

sin(x)cos(y) = [sin(x+y) + sin(x-y)]/2,
cos(x)sin(y) = [sin(x+y) - sin(x-y)]/2,
cos(x)cos(y) = [cos(x-y) + cos(x+y)]/2,
sin(x)sin(y) = [cos(x-y) - cos(x+y)]/2.

sin(x) + sin(y) = 2 sin[(x+y)/2]cos[(x-y)/2],
sin(x) - sin(y) = 2 cos[(x+y)/2]sin[(x-y)/2],
cos(x) + cos(y) = 2 cos[(x+y)/2]cos[(x-y)/2],
cos(x) - cos(y) = -2 sin[(x+y)/2]sin[(x-y)/2],
tan(x) + tan(y) = sin(x+y)/[cos(x)cos(y)],
tan(x) - tan(y) = sin(x-y)/[cos(x)cos(y)],
cot(x) + cot(y) = sin(x+y)/[sin(x)sin(y)],
cot(x) - cot(y) = -sin(x-y)/[sin(x)sin(y)].

[sin(x)+sin(y)]/[cos(x)+cos(y)] = tan[(x+y)/2],
[sin(x)-sin(y)]/[cos(x)+cos(y)] = tan[(x-y)/2],
[sin(x)+sin(y)]/[cos(x)-cos(y)] = -cot[(x-y)/2],
[sin(x)-sin(y)]/[cos(x)-cos(y)] = -cot[(x+y)/2],
[sin(x)+sin(y)]/[sin(x)-sin(y)] = tan[(x+y)/2]/tan[(x-y)/2].

sin2(x) - sin2(y) = sin(x+y)sin(x-y),
cos2(x) - cos2(y) = -sin(x+y)sin(x-y),
cos2(x) - sin2(y) = cos(x+y)cos(x-y).

sin2(x) = (1 - cos[2x])/2,
cos2(x) = (1 + cos[2x])/2,
tan2(x) = (1 - cos[2x])/(1 + cos[2x]),

sin3(x) = (3 sin[x] - sin[3x])/4,
cos3(x) = (3 cos[x] + cos[3x])/4,

sin4(x) = (3 - 4 cos[2x] + cos[4x])/8,
cos4(x) = (3 + 4 cos[2x] + cos[4x])/8,

sin5(x) = (10 sin[x] - 5 sin[3x] + sin[5x])/16,
cos5(x) = (10 cos[x] + 5 cos[3x] + cos[5x])/16,

sin6(x) = (10 - 15 cos[2x] + 6 cos[4x] - cos[6x])/32,
cos6(x) = (10 + 15 cos[2x] + 6 cos[4x] + cos[6x])/32,
```

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### Relations in Right Triangles

In the right triangle ABC with right angle C = /2,

```   A + B = /2,
c2 = a2 + b2,
sin(A) = cos(B) = a/c,
cos(A) = sin(B) = b/c,
tan(A) = cot(B) = a/b,
cot(A) = tan(B) = b/a,
sec(A) = csc(B) = c/b,
csc(A) = sec(B) = c/a,
ha = b,
hb = a,
hc = ab/c.
```

### Solving Right Triangles

Case I: You are given a and A.
B = /2 - A, c = a csc(A), b = a cot(A).

Case II: You are given a and B.
A = /2 - B, c = a sec(B), b = a tan(B).

Case III: You are given c and A.
B = /2 - A, a = c sin(A), b = c cos(A).

Case IV: You are given a and b.
tan(A) = a/b, B = /2 - A, c = a csc(A).

Case V: You are given a and c.
sin(A) = a/c, B = /2 - A, b = a cot(A).

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### Relations in Oblique Triangles

```   A + B + C = ,
s = (a+b+c)/2, half the perimeter,
r = radius of inscribed circle,
R = radius of circumscribed circle,
K = area.
```
The Law of Sines:
```   a/sin(A) = b/sin(B) = c/sin(C) = 2R.
```
This implies that a <= b <= c if and only if A <= B <= C.

The Law of Cosines:

```   a2 = b2 + c2 - 2bc cos(A),
b2 = c2 + a2 - 2ca cos(B),
c2 = a2 + b2 - 2ab cos(C).
```
The Law of Tangents:
```   (a+b)/(a-b) = tan[(A+B)/2]/tan[(A-B)/2],
(b+c)/(b-c) = tan[(B+C)/2]/tan[(B-C)/2],
(c+a)/(c-a) = tan[(C+A)/2]/tan[(C-A)/2].
```
Newton's Formulae:
```   (a+b)/c = cos[(A-B)/2]/sin(C/2),
(b+c)/a = cos[(B-C)/2]/sin(A/2),
(c+a)/b = cos[(C-A)/2]/sin(B/2).
```
Mollweide's Equations:
```   (a-b)/c = sin[(A-B)/2]/cos(C/2),
(b-c)/a = sin[(B-C)/2]/cos(A/2),
(c-a)/b = sin[(C-A)/2]/cos(B/2).
```
Other relations:
```   a = b cos(C) + c cos(B),
b = c cos(A) + a cos(C),
c = a cos(B) + b cos(A).

tan[(A-B)/2] = [(a-b)/(a+b)]cot(C/2),
tan[(B-C)/2] = [(b-c)/(b+c)]cot(A/2),
tan[(C-A)/2] = [(c-a)/(c+a)]cot(B/2).

sin(A) = 2K/(bc),
sin(B) = 2K/(ca),
sin(C) = 2K/(ab).

K = sr = sqrt[s(s-a)(s-b)(s-c)],
K = aha/2 = bhb/2 = chc/2,
K = ab sin(C)/2 = bc sin(A)/2 = ca sin(B)/2,
K = a2 sin(B)sin(C)/[2 sin(A)],
= b2 sin(C)sin(A)/[2 sin(B)],
= c2 sin(A)sin(B)/[2 sin(C)].

R = abc/(4K) = a/[2 sin(A)] = b/[2 sin(B)] = c/[2 sin(C)],
r = K/s,
= sqrt[(s-a)(s-b)(s-c)/s],
= c sin(A/2)sin(B/2)/cos(C/2),
= ab sin(C)/(2s),
= (s-c)tan(C/2).

sin(A/2) = sqrt[(s-b)(s-c)/(bc)],
sin(B/2) = sqrt[(s-c)(s-a)/(ca)],
sin(C/2) = sqrt[(s-a)(s-b)/(ab)].

cos(A/2) = sqrt[s(s-a)/(bc)],
cos(B/2) = sqrt[s(s-b)/(ca)],
cos(C/2) = sqrt[s(s-c)/(ab)].

tan(A/2) = sqrt[(s-b)(s-c)/{s(s-a)}] = r/(s-a),
tan(B/2) = sqrt[(s-c)(s-a)/{s(s-b)}] = r/(s-b),
tan(C/2) = sqrt[(s-a)(s-b)/{s(s-c)}] = r/(s-c).

(a+b)/(a-b) = [sin(A)+sin(B)]/[sin(A)-sin(B)] = cot(C/2)/tan[(A-B)/2],
(b+c)/(b-c) = [sin(B)+sin(C)]/[sin(B)-sin(C)] = cot(A/2)/tan[(B-C)/2],
(c+a)/(c-a) = [sin(C)+sin(A)]/[sin(C)-sin(A)] = cot(B/2)/tan[(C-A)/2].

ha = a sin(B)sin(C)/sin(B+C) = a/[cot(B)+cot(C)] = b sin(C) = c sin(B),
hb = b sin(C)sin(A)/sin(C+A) = b/[cot(C)+cot(A)] = c sin(A) = a sin(C),
hc = c sin(A)sin(B)/sin(A+B) = c/[cot(A)+cot(B)] = a sin(B) = b sin(A).

cos(A) + cos(B) + cos(C) = 1 + r/R.

```
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### Solving Oblique Triangles

Case I: You are given any two angles and one side c.

The third angle is determined from A + B + C = . Now the Law of Sines can be used to find b = c sin(B)/sin(C) and a = c sin(A)/sin(C).

Case II: You are given two sides and the angle opposite one of them, say a, c, and A.

Subcase A: a < c sin(A). There is no solution.

Subcase B: a = c sin(A). There is one solution:
C = /2, B = /2 - A, b = c cos(A).

Subcase C: c > a > c sin(A). There are two solutions. Use the Law of Sines to find sin(C) = c sin(A)/a < 1. There are two angles C and C' = - C having that sine, one acute and one obtuse. Then compute B = - A - C and B' = - A - C'. Now use the Law of Sines again to find b = a sin(B)/sin(A) and b' = a sin(B')/sin(A). The solutions are (a,b,c,A,B,C) and (a,b',c,A,B',C').

Subcase D: a >= c. There is one solution. Use the Law of Sines to find sin(C) = c sin(A)/a <= 1. Then angle C must be acute, so it can be found uniquely from sin(C). Then compute B = - A - C. Now use the Law of Sines again to find b = a sin(B)/sin(A).

Case III: You are given two sides and the included angle, say a, b, and C.

You can compute the third side c by using the Law of Cosines. Then the Law of Sines can be used to find the sines of the other two angles sin(A) = a sin(C)/c and sin(B) = b sin(C)/c. The angles opposite the two shortest sides are then acute, and uniquely determined from their sines, and the third, largest angle is found from A + B + C = .

Alternatively, you can use the Law of Tangents. You know that (A+B)/2 = (-C)/2, which is easily computable. Then by the Law of Tangents, tan[(A-B)/2] = cot(C/2) (a-b)/(a+b), so you can find (A-B)/2 uniquely. Then A = (-C)/2 + (A-B)/2, and B = (-C)/2 - (A-B)/2. Then c = a sin(C)/sin(A).

Case IV: You are given all three sides.

You can use the Law of Cosines to find A, then use the Law of Sines to compute sin(B) = b sin(A)/a and sin(C) = c sin(A)/a.

Alternatively, you can find r = sqrt[(s-a)(s-b)(s-c)/s], and use tan(A/2) = r/(s-a) to find A/2, and hence A, and similarly for B and C.

Alternatively, you can use sin(A/2) = sqrt[(s-b)(s-c)/(bc)] to find A/2 (since A/2 < /2), and hence A, and similarly for B and C.

In any case, the results can be checked by using Mollweide's Equations.

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### Inverse Trigonometric Functions

```   x = Arcsin(y)  ==>  y = sin(x),  -/2 <= x <= /2,
x = Arccos(y)  ==>  y = cos(x),     0 <= x <= ,
x = Arctan(y)  ==>  y = tan(x),  -/2 <  x <  /2,
x = Arccot(y)  ==>  y = cot(x),     0 <  x <  ,
x = Arcsec(y)  ==>  y = sec(x),     0 <  x <  ,
x = Arccsc(y)  ==>  y = csc(x),  -/2 <  x <  /2.

y = sin(x)  ==>  x = Arcsin(y) + 2n  or   - Arcsin(y) + 2n,
y = cos(x)  ==>  x = Arccos(y) + 2n  or  -Arccos(y)    + 2n,
y = tan(x)  ==>  x = Arctan(y) +  n,
y = cot(x)  ==>  x = Arccot(y) +  n,
y = sec(x)  ==>  x = Arcsec(y) + 2n  or  -Arcsec(y)    + 2n,
y = csc(x)  ==>  x = Arccsc(y) + 2n  or   - Arccsc(y) + 2n,
```
where n is an arbitrary integer.
```   Arcsin(y) + Arccos(y) = /2,
Arctan(y) + Arccot(y) = /2,
Arcsec(y) + Arccsc(y) = /2.

Arcsin(y) = Arccsc(1/y),
Arccos(y) = Arcsec(1/y),
Arctan(y) = Arccot(1/y),  y > 0
Arccot(y) = Arctan(1/y),  y > 0
Arcsec(y) = Arccos(1/y),
Arccsc(y) = Arcsin(1/y).

Arcsin[sin(x)] = (-1)Floor[1/2 - x/] (x +  Floor[1/2 - x/]),
Arccos[cos(x)] = /2 - (-1)Floor[-x/] (x + /2 +  Floor[-x/])
Arctan[tan(x)] = x - Floor[x/ + 1/2] .

sin[Arcsin(y)] = cos[Arccos(y)] = y,
cos[Arcsin(y)] = sin[Arccos(y)] = sqrt[1-y2],
tan[Arcsin(y)] = cot[Arccos(y)] = y/sqrt[1-y2],
cot[Arcsin(y)] = tan[Arccos(y)] = sqrt[1-y2]/y,
sec[Arcsin(y)] = csc[Arccos(y)] = 1/sqrt[1-y2],
csc[Arcsin(y)] = sec[Arccos(y)] = 1/y,
sin[Arctan(y)] = cos[Arccot(y)] = y/sqrt[1+y2],
cos[Arctan(y)] = sin[Arccot(y)] = 1/sqrt[1+y2],
tan[Arctan(y)] = cot[Arccot(y)] = y,
cot[Arctan(y)] = tan[Arccot(y)] = 1/y,
sec[Arctan(y)] = csc[Arccot(y)] = sqrt[1+y2],
csc[Arctan(y)] = sec[Arccot(y)] = sqrt[1+y2]/y,
sin[Arcsec(y)] = cos[Arccsc(y)] = sqrt[y2-1]/y,
cos[Arcsec(y)] = sin[Arccsc(y)] = 1/y,
tan[Arcsec(y)] = cot[Arccsc(y)] = sqrt[y2-1],
cot[Arcsec(y)] = tan[Arccsc(y)] = 1/sqrt[y2-1],
sec[Arcsec(y)] = csc[Arccsc(y)] = y,
csc[Arcsec(y)] = sec[Arccsc(y)] = y/sqrt[y2-1].

sin[2 Arcsin(y)] = 2y sqrt[1-y2],
cos[2 Arccos(y)] = 2y2 - 1,
tan[2 Arctan(y)] = 2y/(1-y2).
```

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Compiled by Robert L. Ward.