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
More values at special angles:
x
Use the above values and the identities below to obtain values of trigonometric functions of the following multiples of /10:
3
Bounds
|sin(x)| <= 1, |cos(x)| <= 1, |sec(x)| >= 1, |csc(x)| >= 1.
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(
Relations in Right Triangles
In the right triangle ABC with right angle C = /2,
A + B =
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).
Relations in Oblique Triangles
A + B + C =
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.
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’ = having that sine, one acute and one obtuse. Then compute – CB = and – A – CB’ = Now use the Law of Sines again to find – A – C’.b = a sin(B)/sin(A) andb’ = 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 andsin(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 fromA + B + C = . - Alternatively, you can use the Law of Tangents. You know that
(A+B)/2 = ( which is easily computable. Then by the Law of Tangents,-C)/2,tan[(A-B)/2] = cot(C/2) (a-b)/(a+b), so you can find (A-B)/2 uniquely. ThenA = ( and-C)/2 + (A-B)/2,B = ( Then-C)/2 – (A-B)/2.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 andsin(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 forB and C. - Alternatively, you can use sin(A/2) = sqrt[(s-b)(s-c)/(bc)] to find A/2 (since
A/2 < and hence A, and similarly for/2),B and C.
In any case, the results can be checked by using Mollweide’s Equations.
Inverse Trigonometric Functions
x = Arcsin(y) ==> y = sin(x), -
where n is an arbitrary integer.
Arcsin(y) + Arccos(y) =
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