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Lesson Note On Definitions of Trigonometric Functions and Its Equations

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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.


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(/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,


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).

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.


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 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 = (-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. ThenA = (-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 computesin(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.


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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