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Mathematics Dictionary
Dr. K. G. Shih
Trigonometric Functions of Compound Angles
Subjects


  • TR 07 00 | - Outlines
  • TR 07 01 | - sin(A+B)=sin(A)*cos(B)+cos(A)*sin(B)
  • TR 07 02 | - cos(A+B)=cos(A)*cos(B)-sin(A)*sin(B)
  • TR 07 03 | - sin(A-B)=sin(A)*cos(B)-cos(A)*sin(B)
  • TR 07 04 | - cos(A-B)=cos(A)*cos(B)+sin(A)*sin(B)
  • TR 07 05 | - tan(A+B) and Tan(A-B)
  • TR 07 06 | - Multiple angles : sin(5*A)
  • TR 07 07 | - Half angle
  • TR 07 08 | - Product of functions to sum of functions
  • TR 07 09 | - Sum of functions to product of functions
  • TR 07 10 | - sin(A+B)*sin(A-B) = sin(A)^2 - sin(B)^2
  • TR 07 11 | - Angles of triangle in 3 GP terms with common ratio = 3
  • TR 07 12 | - (sin(A) + sin(B))^2 = 1 + sin(2*A)
  • TR 07 13 | - Prove that cos(6*x) = 32*cos(x)^6 - 48*cos(x)^4 +18*cos(x)^2 - 1
  • TR 07 14 | - S(n) = cos(x) + cos(3*x) + cos(5*x) + .... + cos((2*n-1)*x)
  • TR 07 15 | - S(n) = sin(x)^2 + sin(2*x)^2 + .... + sin(n*x)^2
  • TR 07 16 | - Angles of triangle in 3 GP terms with common ratio = 2
  • TR 07 17 | - y = sin(x) + cos(x), find maximum
  • TR 07 18 | - y = sin(x) + Sqr(3)*cos(x), find maximum
  • TR 07 19 | - Triangle ABC : Identities
  • TR 07 20 | - sin(A)+sin(B)=p and cos(A)+cos(B)=q, find sin(A+B) and cos(A+B)
  • TR 07 21 | - sin(A) and cos(A) have AP mid term sin(x) and GP mid term sin(y)
  • TR 07 22 | - If sin(x+y) = sin(x)+sin(y), ....
  • TR 07 23 | - Show that 5 + 8*cos(x) + 4*cos(2*x) + cos(3*x) GE 0
  • TR 07 24 | - Find sin(20)*sin(40)*sin(60)*sin(80) with out calculator
  • TR 07 25 | - sin(pi/16)^4 + sin(3*pi/16)^4 + sin(5*pi/16)^4 + sin(7*pi/16) = ?
  • TR 07 26 | - Quiz and answers
    • Answers


    TR 07 01. sin(A+B) = sin(A)*cos(B) + cos(A)*sin(B)

    Prove that sin(A+B) = sin(A)*cos(B) + cos(A)*sin(B).
    • Figure 10-01
    • PN perpendicular to ON and NK perpemdicular to PL.
    • Right triangle ONP.
      • NP = OP*sin(B) and NO = OP*cos(B).
    • Right triangle OPL
      • sin(A+B) = PL/OP = (PK + KL)/OP = (PK + MN)/OP.
      • PK = NP*cos(A) = OP*sin(B)*cos(A).
      • MN = ON*sin(A) = OP*cos(B)*sin(A).
      • sin(A+B) = sin(A)*cos(B) + cos(A)*sin(B).
    Application of sin(A+B) : Find values
    • sin(075) = sin(30+ 45) = sin(30)*c0s(45) + cos(30)*sin(45) = ?
    • sin(105) = sin(180-75) = +sin(75)
    • sin(255) = sin(180+75) = -sin(75)
    • sin(285) = sin(360-75) = -sin(75)
    Application of sin(A+B) : Find drivative of sin(x)
    • Lim[sin(x+h) - sin(x))/h] = cos(x) as h tends to 0.

    Go to Begin

    TR 07 02. cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B)

    Prove that cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B).
    • Figure 10-01
    • PN perpendicular to ON and NK perpemdicular to PL.
    • Right triangle ONP
      • NP = OP*sin(B) and NO = OP*cos(B).
    • Right triangle OPL
      • cos(A+B) = OL/OP = (OM - LM)/OP = (OM - KN)/OP.
      • KN = NP*sin(A) = OP*sin(B)*sin(A).
      • OM = ON*cos(A) = OP*cos(B)*cos(A).
      • cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B).
    Application of cos(A+B)
    • cos(075) = cos(30+ 45) = cos(30)*c0s(45) - cos(30)*sin(45) = ?
    • cos(105) = cos(180-75) = -cos(75)
    • cos(255) = cos(180+75) = -cos(75)
    • cos(285) = cos(360-75) = +cos(75)

    Go to Begin

    TR 07 03. sin(A-B) = sin(A)*cos(B) - cos(A)*sin(B)

    Prove that sin(A-B) = sin(A)*cos(B) + cos(A)*sin(B).
    • Figure 10-02
    • PN perpendicular to ON and NK perpemdicular to PL.
    • Right triangle ONP.
      • NP = OP*sin(B) and NO = OP*cos(B).
    • Right triangle OPL
      • sin(A-B) = PL/OP = (KL - PK)/OP = (MN - OK)/OP.
      • PK = NP*cos(A) = OP*sin(B)*cos(A).
      • MN = ON*sin(A) = OP*cos(B)*sin(A).
      • sin(A-B) = sin(A)*cos(B) - cos(A)*sin(B).
    Application of sin(A-B)
    • sin(015) = sin(45- 30) = sin(30)*c0s(45) + cos(30)*sin(45) = ?
    • sin(165) = sin(180-15) = +sin(15)
    • sin(195) = sin(180+15) = -sin(15)
    • sin(345) = sin(360-15) = -sin(15)

    Go to Begin

    TR 07 04. cos(A-B) = cos(A)*cos(B) + sin(A)*sin(B)
    Geometric method
    • Figure 10-02
    • PN perpendicular to ON and NK perpemdicular to PL.
    • Right triangle ONP
      • NP = OP*sin(B) and NO = OP*cos(B).
    • Right triangle OPL
      • cos(A-B) = OL/OP = (OM + LM)/OP = (OM + KN)/OP.
      • KN = NP*sin(A) = OP*sin(B)*sin(A).
      • OM = ON*cos(A) = OP*cos(B)*cos(A).
      • cos(A+B) = cos(A)*cos(B) + sin(A)*sin(B).
    Use cosine law and distance formuls
    • Construction
      • Draw a unit circle with OX = 1.
      • Draw point P and Q on circle. OP = OQ = 1.
      • Angle XOP = A and XOQ = B
      • Angle POQ = A - B
      • Distance = AB = d
    • Proof
      • By cosine law
        • d^2 = 1^2 + 1^2 - 2*1*1*cos(A-B)
        • d^2 = 2 - 2*cos(A-B)
      • Find distance between P and Q
        • Coordinate P(cos(A), sin(A))
        • Coordinate Q(cos(B), sin(B))
        • d^2 = (cos(A)-cos(B))^2 + (sin(A)-sin(B))^2
        • Since cos(A)^2 + sin(A)^2 = 1 and cos(B)^2 + sin(B)^2 = 1
        • After expansion
        • d^2 = 2 - 2*cos(A)*cos(B) - 2*sin(A)*sin(B)
      • Hence cos(A-B) = cos(A)*cos(B) + sin(A)*sin(B)
    Application of cos(A-B)
    • cos(015) = cos(45- 30) = cos(30)*cos(45) + sin(30)*sin(45) = ?
    • cos(165) = cos(180-15) = -cos(15)
    • cos(195) = cos(180+15) = -cos(15)
    • cos(345) = cos(360-15) = +cos(15)

    Go to Begin

    TR 07 05. tan(A+B) and tan(A-B)

    Prove that tan(x+y) = (tan(x) + tan(y))/(1 - tan(x)*tan(y)).
    • tan(A+B) = (sin(A)*cos(B) + cos(A)*sin(B))/(sin(A)*cos(B) os(A)*cos(B)).
    • Divide numerator and denominator by cos(A)*cos(B).
    • tan(A+B) = (sin(A)/cos(A) + sin(B)/cos(B))/(1 - (sin(A)/cos(A))*(sin(B)/cos(B))).
    • tan(A+B) = (tan(A) +tan(B))/(1 - tan(A)*tan(B)
    Prove that tan(x-y) = (tan(x) - tan(y))/(1 + tan(x)*tan(y)).
    • tan(A-B) = (sin(A)*cos(B) - cos(A)*sin(B))/(sin(A)*cos(B) + cos(A)*cos(B)).
    • Divide numerator and denominator by cos(A)*cos(B).
    • tan(A-B) = (sin(A)/cos(A) - sin(B)/cos(B))/(1 + (sin(A)/cos(A))*(sin(B)/cos(B))).
    • tan(A-B) = (tan(A) - tan(B))/(1 + tan(A)*tan(B)
    Application of tan(A+B)
    • tan(075) = tan(30+ 45) = (tan(30) + tan(45))/(1 - tan(30)*tan(45)) = ?
    • tan(105) = tan(180-75) = -tan(75)
    • tan(255) = tan(180+75) = +tan(75)
    • tan(285) = tan(360-75) = -tan(75)
    Application of tan(A-B)
    • tan(015) = tan(45- 30) = (tan(45) - tan(30))/(1 + tan(30)*tan(45)) = ?
    • tan(165) = tan(180-15) = -tan(15)
    • tan(195) = tan(180+15) = +tan(15)
    • tan(345) = tan(360-15) = -tan(15)

    Go to Begin

    TR 07 06. Multple of angle : sin(5*x)

    Identities of sine :
    • 1. sin(2*x) = 2*sin(x)*cos(x).
    • 2. sin(3*x) = 3*sin(x) - 4*sin(x)^3.
    • 3. sin(4*x) = 4*sin(x)*cos(x)^3 - 4*cos(x)*sin(x)^3.
    • 4. sin(5*x) = 5*sin(x) - 20*sin(x)^3 + 16*sin(x)^5.
    Identities of cosine
    • 1. cos(2*x) = cos(x)^2 - sin(x)^2 = 2*cos(x)^2 - 1 = 1 - 2*sin(x)^2.
    • 2. cos(3*x) = 4*cos(x)^3 - 3*cos(x).
    • 3. cos(4*x) = 8*cos(x)^4 - 8*cos(x)^2 + 1.
    • 4. cos(5*x) = 16*cos(x)^5 - 20*cos(x)^3 + 5*cos(x)
    Identities of tangent
    • 1. tan(2*x) = 2*tan(x)/(1-tan(x)^2).

    Go to Begin

    TR 07 07. Half angle

    Identities
    • sin(A/2) = Sqr((1-cos(A))/2)
    • cos(A/2) = Sqr((1+cos(A))/2)
    • tan(A/2) = Sqr((1-cos(A))/(1+cos(A)))
    Proof
    • Prove that sin(A/2) = Sqr((1-cos(A))/2).
      • Since cos(2*x) = 1 - 2*sin(x)^2.
      • Hence 2*sin(x)^2 = 1 - cos(2*x).
      • Let 2*x = A we have sin(A/2)^2 = 1 - cos(A).
      • Hence sin(A/2) = Sqr((1-cos(A))/2).
    • Prove that cos(A/2) = Sqr((1+cos(A))/2).
      • Since cos(2*x) = 2*cos(x)^2 - 1.
      • Hence 2*cos(x)^2 = 1 + cos(2*x).
      • Let 2*x = A we have cos(A/2)^2 = 1 + cos(A).
      • Hence cos(A/2) = Sqr((1+cos(A))/2).
    Application examples
    • 1. Find sin(22.5) without using calculator.
      • sin(22.5) = Sqr((1 - cos(45))/2).
      • Since cos(45) = Sqr(2)/2.
      • Hence sin(22.5) = Sqr((1 - Sqr(2)/2)/2).
    • 2. Find sin(15) without using calculator.
      • sin(15) = Sqr((1 - cos(30))/2).
      • Since cos(30) = Sqr(3)/2.
      • Hence sin(15) = Sqr((1 - Sqr(3)/2)/2).
    • 3. Find cos(22.5)
      • cos(22.5) = Sqr((1+cos(45))/2)
      • = Sqr((1+Sqr(2))/2)/2)
      • = Sqr(2+Sqr(2))/4).

    Go to Begin

    TR 07 08. Product of functions to sum of functions

    Identities : Use sum and difference we can easily derive the formula
    • sin(A)*sin(B) = (cos(A-B) - cos(A+B))/2.
    • cos(A)*cos(B) = (cos(A-B) + cos(A+B))/2.
    • sin(A)*cos(B) = (sin(A+B) - sin(A-B))/2.
    • cos(A)*sin(A) = (cos(A+B) + cos(A-B))/2.
    Proof
    • 1. Prove that sin(A)*sin(B) = (cos(A-B) - cos(A+B))/2.
      • cos(A+B) = cos(A)*cos(B) - sin(A)*(sin(B).
      • cos(A-B) = cos(A)*cos(B) + sin(A)*(sin(B).
      • Hence 2*sin(A)*sin(B) = cos(A-B) - cos(A+B)
    • 2. Prove that cos(A)*cos(B) = (cos(A-B) + cos(A+B))/2.
      • cos(A+B) = cos(A)*cos(B) - sin(A)*(sin(B).
      • cos(A-B) = cos(A)*cos(B) + sin(A)*(sin(B).
      • Hence 2*cos(A)*cos(B) = cos(A+B) + cos(A-B)

    Go to Begin

    TR 07 09. Sum of functions to Product functions

    Identities : Use product of functions to derive the following formula
    • sin(A) + sin(B) = +2*sin((A+B)/2)*cos((A-B)/2).
    • sin(A) - sin(B) = +2*cos((A+B)/2)*sin((A-B)/2).
    • cos(A) + cos(B) = +2*cos((A+B)/2)*cos((A-B)/2).
    • cos(A) - cos(B) = -2*sin((A+B)/2)*sin((A-B)/2).
    Proof
    • 1. Prove that sin(A) + sin(B) = 2*sin((A+B)/2)*cos((A-B)/2).
      • sin(C+D) = sin(C)*cos(D) + cos(C)*sin(D).
      • sin(C-D) = sin(C)*cos(D) - cos(C)*sin(D).
      • sin(C+D) + sin(C-D) = 2*sin(C)*cos(D).
      • Let C + D = A and C - D = B.
      • Hence 2*C = A + B and 2*D = A - B.
      • Hence sin(A) + sin(B) = 2*sin((A+B)/2)*cos((A-B)/2).
    • 2. Prove that sin(A) - sin(B) = 2*cos((A+B)/2)*sin((A-B)/2).
      • sin(C+D) = sin(C)*cos(D) + cos(C)*sin(D).
      • sin(C-D) = sin(C)*cos(D) - cos(C)*sin(D).
      • sin(C+D) - sin(C-D) = 2*cso(C)*sin(D).
      • Let C + D = A and C - D = B.
      • Hence 2*C = A + B and 2*D = A - B.
      • Hence sin(A) - sin(B) = 2*cos((A+B)/2)*sin((A-B)/2).

    Go to Begin

    TR 07 10. sin(A+B)*sin(A-B) = sin(A)^2 - sin(B)^2

    Hint
    • 1. Formula of sin(A+B) and sin(A-B).
    • 2. Factor : (x^2 - y^2) = (x + y)*(x - y).
    • 3. Pythagorean relation : cos(x)^2 + sin(x)^2 = 1.
    Proof
    • LHS = (sin(A)*cos(B)+cos(A)*sin(B)*(sin(A)*cos(B)-cos(A)*sin(B).
    • LHS = (sin(A)*cos(B))^2 - (cos(A)*sin(B))^2.
    • LHS = (sin(A)^2)*(1-sin(B)^2) - (sin(B)^2)*(1-sin(B)^2).
    • LHS = sin(A)^2 - sin(B)^2.
    • LHS = RHS.

    Go to Begin

    TR 07 11. Angles of triangle in 3 GP terms with common ratio = 3

    Question
    • Angles of triangle are in 3 consecutive GP terms.
    • If common ratio is r = 3, then cos(A)*cos(B) + cos(B)*cos(C) + cos(C)*cos(A) = -1/4
    Proof
    • Let angle A = x, then angle B = 3*x and angle C = 9*x
    • A + B + C = 13*x = 180. Hence angle A = x = 180/13 degrees
    • Use product formula of functions
    • LHS
      • = cos(A)*cos(B) + cos(B)*cos(C) + cos(C)*cos(A)
      • = (cos(A+B) + cos(A-B) + cos(B+C) + cos(B-C) + cos(C+A) + cos(C-A))/2
      • = (cos(4*x) + cos(2*x) + cos(12*x) + cos(6*x) + cos(10*x) + cos(8*x))/2
      • = (cos(12*x) + cos(10*x) + cos(8*x) + cos(6*x) + cos(4*x) + cos(2*x))/2
    • Each term of above multiply by sin(x)/sin(x)
    • LHS = (cos(12*x)*sin(x) + cos(10*x)*sin(x) + .... )/(2*sin(x))
    • Since
      • sin(x)*cos(2*x) = (sin(x+2*x) + sin(x-2*x))/2 = (sin(3*x) - sin(x))/2
      • sin(x)*cos(4*x) = (sin(x+4*x) + sin(x-4*x))/2 = (sin(5*x) - sin(3*x))/2
      • sin(x)*cos(6*x) = (sin(x+6*x) + sin(x-6*x))/2 = (sin(7*x) - sin(5*x))/2
      • sin(x)*cos(8*x) = (sin(x+8*x) + sin(x-8*x))/2 = (sin(9*x) - sin(7*x))/2
      • sin(x)*cos(10*x) = (sin(x+10*x) + sin(x-10*x))/2 = (sin(11*x) - sin(9*x))/2
      • sin(x)*cos(12*x) = (sin(x+12*x) + sin(x-12*x))/2 = (sin(13*x) - sin(11*x))/2
    • Add above terms and many terms are cancelled out
    • LHS = (sin(13*x) - sin(x))/(2*2*sin(x)
    • LHS = (sin(180) - sin(x))/(4*sin(x))
    • LHS = (0 - sin(x))/(4*sin(x))
    • LHS = -1/4
    Exercise : Numerical proof
    • Since A = x, B = 3*x and C = 9*x are in GP with common ratio = 3
    • Hence A + B + C = 13*x = 180 degree
    • Hence A = x = 180/13 = 13.84615, B = 3*x = 41.53846 and C = 9*x = 124.61538
    • Substitute A, B, C into expression
      • cos(A)*cos(B) + cos(B)*cos(C) + cos(C)*cos(A) = ?
      • Use calculator find the answer

    Go to Begin

    TR 07 12. Prove (sin(A) + cos(A))^2 = 1 + sin(2*A)

    Proof
    • (sin(A) + cos(A))^2 = sin(A)^2 + 2*sin(A)*cos(A) + cos(A)^2
    • = (cos(A)^2 + sin(A)^2) + 2*sin(A)*cos(A)
    • = 1 + sin((2*A).

    Go to Begin

    TR 07 13. Prove that cos(6*x) = 32*cos(x)^6 - 48*cos(x)^4 +18*cos(x)^2 - 1

    Proof
    • cos(6*x) = cos(3x + 3*x).
    • cos(3*x) = 2*cos(3*x)^2 - 1.
    • cos(3*x) = 2*(4*cos(x)^3 - 3*cos(x))^2 - 1.
    • cos(6*x) = 32*cos(x)^6 - 48*cos(x)^4 + 18*cos(x)^2 - 1.

    Go to Begin

    TR 07 14. S(n) = cos(x) + cos(3*x) + cos(5*x) + .... + cos((2*n-1)*x)

    Solution
    Go to Begin

    TR 07 15. S(n) = sin(x)^2 + sin(2*x)^2 + sin(3*x)^2 + ... + sin(n*x)^2

    Solution
    Go to Begin

    TR 07 16. Angles of triangle in 3 GP terms with common ratio = 2

    Solution
    Go to Begin

    TR 07 17. y = sin(x) + cos(x), find maximum

    Graphic     Mehtod 1 : Use sin(A+B) = sin(A)*cos(B) - cos(A)*sin(B)
    • Multiply each side by Sqr(2)/2
      • (Sqr(2)*y/2) = (Sqr(2)/2)*sin(x) + (Sqr(2)/2)*cos(x)
      • = cos(pi/4)*sin(x) + sin(pi/4)*cos(x)
      • = sin((pi/4) + x)
    • Since sin(A) LE 1 or GE - 1
    • sin(x + pi/4) = 1 has maximum
    • Hence x + pi/4 = pi/2 or x = pi/4
    • Hence maximum of y is y = Sqr(2)
    Method 2 : Use dy/dx = 0
    • y' = cos(x) - sin(x) = 0
    • Hence cos(x) = sin(x) or tan(x) = 1
    • Hence x = pi/4 and y = Sqr(2)

    Go to Begin

    TR 07 18. y = sin(x) + Sqr(3)*cos(x), find maximum

    Graphic     Mehtod 1 : Use sin(A+B) = sin(A)*cos(B) - cos(A)*sin(B)
    • Multiply each side by 1/2
      • (y/2) = (1/2)*sin(x) + (Sqr(3)/2)*cos(x)
      • = cos(pi/3)*sin(x) + sin(pi/3)*cos(x)
      • = sin((pi/3) + x)
    • Since sin(A) LE 1 or GE - 1
    • sin(x + pi/3) = 1 has maximum
    • Hence x + pi/3 = pi/2 or x = pi/6
    • Hence maximum of y is y = sin(pi/6) + Sqr(3)*cos(pi/6) = 2
    Method 2 : Use dy/dx = 0
    • y' = cos(x) - Sqr(3)*sin(x) = 0
    • Hence cos(x) = Sqr(3)*sin(x) or tan(x) = 1/Sqr(3)
    • Hence x = pi/6 and y = 2

    Go to Begin

    TR 07 19. Triangle ABC : Identities

    Questions
    • 1. cos(2*A) + cos(2*B) + cos(2*C) = ?
    • 2. sin(A)^2 + sin(B)^2 + sin(C)^2 = ?
    • 3. cos(A) + cos(B) + cos(C) = ?
    Solution
    Go to Begin

    TR 07 20. sin(A)+sin(B)=p and cos(A)+cos(B)=q, find sin(A+B) and cos(A+B)

    Solution
    Go to Begin

    TR 07 21. sin(A) and cos(A) have AP mid term sin(x) and GP mid term sin(y), ...

    Solution
    Go to Begin

    TR 07 22. If sin(x + y) = sin(x) + sin(y)

    Solution
    Go to Begin

    TR 07 23. 5 + 8*cos(x) + 4*cos(2*x) + cos(3*x) GE 0

    Solution
    Go to Begin

    TR 07 24. Find sin(20)*sin(40)*sin(60)*sin(80) with out calculator

    Solution
    • Subject | Q09, Q10 and Q11
    • Q09 : Find sin(20)*sin(40)*sin(60)*sin(80) with out calculator
    • Q10 : Find cos(20)*cos(40)*cos(60)*cos(80) with out calculator
    • Q11 : Find cos(20)*cos(40)*cos(60)*cos(80) with out calculator

    Go to Begin

    TR 07 25. sin(pi/16)^4 + sin(3*pi/16)^4 + sin(5*pi/16)^4 + sin(7*pi/16) = 3/2

    Solution
    • Subject | Q17
    • Q18 : cos(1*pi/16)^4 + cos(3*pi/16)^4 + cos(5*pi/16)^4 + cos(7*pi/16)^4 = 3/2

    Go to Begin

    TR 07 26. Quiz and answer

    Answer
    Go to Begin

    Q00. Outlines

    Sum of angles
    • 1. sin(x+y) = sin(x)*cos(y) + cos(x)*sin(y).
    • 2. cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y).
    • 3. tan(x+y) = (tan(x) + tan(y))/(1 - tan(x)*tan(y)).
    Diference of angles
    • 1. sin(x-y) = sin(x)*cos(y) - cos(x)*sin(y).
    • 2. cos(x-y) = cos(x)*cos(y) + sin(x)*(sin(y).
    • 3. tan(x-y) = (tan(x) - tan(y))/(1 + tan(x)*tan(y)).
    Multiple angles
    • cos(2*x) = 2*cos(x)^2 - 1
    • sin(2*x) = 2*sin(x)*cos(x)
    • tan(2*x) = 2*tan(x)/(1-tan(x)^2)
    Half angle
    • cos(x/2) = Sqr((1 + cos(x))/2)
    • sin(x/2) = Sqr((1 - cos(x))/2)
    • cos(x) = Sqr((1 - cos(x))/(1 + cos(x)))
    Product of functions
    • sin(A)*sin(B) = (cos(A-B) - cos(A+B))/2.
    • cos(A)*cos(B) = (cos(A-B) + cos(A+B))/2.
    • sin(A)*cos(B) = (sin(A+B) - sin(A-B))/2.
    • cos(A)*sin(A) = (cos(A+B) + cos(A-B))/2.
    Sum or difference of functions
    • sin(A) + sin(B) = +2*sin((A+B)/2)*cos((A-B)/2).
    • sin(A) - sin(B) = +2*cos((A+B)/2)*sin((A-B)/2).
    • cos(A) + cos(B) = +2*cos((A+B)/2)*cos((A-B)/2).
    • cos(A) - cos(B) = -2*sin((A+B)/2)*sin((A-B)/2).

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