Having a lot of fun working with my kid to proof the trigonometric formulas in the camp textbook.
Sum and Difference Formulas
(1) cos(u-v) = cos(u)cos(v) + sin(u)sin(v)
Let A, B, C and D are points on unit circle O. A is [1,0], B is [cos(u-v), sin(u-v)], C is [cos(v), sin(v)] and D is [cos(u), sin(u)]. Then angle AOB is equals to angle COD, hence, segment AB = CD.
AB2 = [cos(u-v)-1]2 + [sin(u-v)-0]2 = CD2 =[(cos(u)-cos(v)]2 + [sin(u) – sin(v)]2 ------ Distance formula.
=> [cos(u-v)]2 – 2cos(u-v) + 1 + [sin(u-v)]2 = [(cos(u)]2 – 2cos(u)cos(v) + [cos(v)]2 + [sin(u)]2 – 2sin(u)sin(v) + [sin(v)]2
=> – 2cos(u-v) + 2 = 2 – 2cos(u)cos(v) – 2sin(u)sin(v) ----- Using [(cos(x)]2 +[sin(x)]2 = 1
=> cos(u-v) = cos(u)cos(v) + sin(u)sin(v)
(2) cos(u+v) = cos(u)cos(v) - sin(u)sin(v)
cos(u+v) = cos(u-(-v)) = cos(u)cos(-v) + sin(u)sin(-v) ------ Using (1)
=> cos(u+v) = cos(u)cos(v) – sin(u)(sin(v) ------- Using cos(-x) = cos(x) and sin(-x) = –sin(x)
(3) sin(u+v) = sin(u)cos(v) + cos(u)sin(v)
sin(u+v) = cos(pi/2 – (u+v)) = cos((pi/2-u) – v) = cos(pi/2-u)cos(v) + sin(pi/2-u)sin(v) ------- Using (1)
=> sin(u+v) = sin(u)cos(v) + cos(u)sin(v) ------ Using cos(pi/2-x) = sin(x) and sin(pi/2-x) = cos(x)
(4) sin(u-v) = sin(u)cos(v) - cos(u)sin(v)
sin(u-v) = sin(u+(-v)) = sin(u)cos(-v) + cos(u)sin(-v) ------ Using (3)
=> sin(u-v) = sin(u)cos(v) – cos(u)(sin(v) ------- Using cos(-x) = cos(x) and sin(-x) = –sin(x)
Product-to-Sum Formulas
(5) cos(u)cos(v) = [cos(u+v) + cos(u-v)] / 2
Add formula (1) and (2):
cos(u+v) + cos(u-v) = cos(u)cos(v) - sin(u)sin(v) + cos(u)cos(v) + sin(u)sin(v)
=> cos(u+v) + cos(u-v) = 2cos(u)cos(v)
=> cos(u)cos(v) = [cos(u+v) + cos(u-v)] / 2
(6) sin(u)sin(v) = [cos(u-v) - cos(u+v)] / 2
Subtract formula (1) by (2):
cos(u-v) - cos(u+v) = cos(u)cos(v) + sin(u)sin(v) – [cos(u)cos(v) - sin(u)sin(v)]
=> cos(u-v) - cos(u+v) = 2sin(u)sin(v)
=> sin(u)sin(v) = [cos(u-v) - cos(u+v)] / 2
(7) sin(u)cos(v) = [sin(u+v) + sin(u-v)] / 2
Add formula (3) and (4):
sin(u+v) + sin(u-v) = sin(u)cos(v) + cos(u)sin(v) + sin(u)cos(v) - cos(u)sin(v)
=> sin(u+v) + sin(u-v) = 2sin(u)cos(v)
=> sin(u)cons(v) = [sin(u+v) + sin(u-v)] / 2
(8) cos(u)sin(v) = [sin(u+v) - sin(u-v)] / 2
There are two simple methods to prove this.
Method 1
Subtract formula (3) by (4)
sin(u+v) - sin(u-v) = sin(u)cos(v) + cos(u)sin(v) – [sin(u)cos(v) - cos(u)sin(v)]
=> sin(u+v) + sin(u-v) = cos(u)sin(v)
=> cos(u)sin(v) = [sin(u+v) - sin(u-v)] / 2
Method 2
cos(u)sin(v) = sin(v)cos(u) = [sin(v+u) + sin(v-u)] / 2 ------ Using (7)
=> cos(u)sin(v) = sin(v)cos(u) = [sin(u+v) + sin(-(u-v))] / 2
=> cos(u)sin(v) = [sin(u+v) - sin(u-v)] / 2 ------ Using sin(-x) = -sin(x)
Sum-to-Product Formulas
(9) cos(u) + cos(v) = 2cos[(u+v)/2]cos[(u-v)/2]
cos(x)cos(y) = [cos(x+y) + cos(x-y)] / 2 ------ Using (5)
Let u=x+y and v=x-y, we can have x=(u+v)/2 and y=(u-v)/2
=> cos[(u+v)/2]cos[(u-v)/2] = [cos(u) + cos(v)] / 2 ------ Replacing x, y with u, v
=> cos(u) + cos(v) = 2cos[(u+v)/2]cos[(u-v)/2]
(10) cos(u) - cos(v) = -2sin[(u+v)/2]sin[(u-v)/2]
Similarly to the proof of (9), we can arrive (10) from (6)
(11) sin(u) + sin(v) = 2sin[(u+v)/2]cos[(u-v)/2]
Similarly to the proof of (9), we can arrive (11) from (7)
(12) sin(u) - sin(v) = 2cos[(u+v)/2]sin[(u-v)/2]
There are two methods to do this.
Method 1
Similarly to the proof of (9), we can arrive (12) from (8)
Method 2
sin(u) – sin(v) = sin(u) + sin(-v) ------ Using sin(-x) = -sin(x)
= 2sin[(u+(-v))/2]cos[(u-(-v))/2] ------ Using (12)
=> sin(u) - sin(v) = 2cos[(u+v)/2]sin[(u-v)/2]
Double Angle Formulas
(13.1) cos(2u) = cos2(u) – sin2(u)
cos(2u) = cos(u+u) = cos(u)cos(u) – sin(u)sin(u) ------ Using (2)
=> cos(2u) = cos2(u) – sin2(u)
(13.2) cos(2u) = 2cos2(u) – 1
cos(2u) = cos2(u) – sin2(u) ------ Using (13.1)
= cos2(u) – [sin2(u) + cos2(u) - cos2(u)]
= cos2(u) – [1-cos2(u)] ------ Using cos2(u) + sin2(u) = 1
= 2cos2(u) – 1
(13.3) cos(2u) = 1 – 2sin2(u)
cos(2u) = cos2(u) – sin2(u) ------ Using (13.1)
= [cos2(u) + sin2(u) - sin2(u)] – sin2(u)
= [1 – sin2(u)] – sin2(u) ------ Using cos2(u) + sin2(u) = 1
= 1 - 2sin2(u)
(14) sin(2u) = sin(u)cos(u)
sin(2u) = sin(u+u) = sin(u)cos(u) + cos(u)sin(u) ------ Using (3)
=> sin(2u) = 2sin(u)cos(u)
(15) tan(2u) = 2tan(u) / (1 - tan2(u))
Divide formula (14) by (13.1)
sin(2u)/cos(2u) = 2sin(u)cos(u)/[cos2(u) – sin2(u)]
=> tan(2u) = 2sin(u)cos(u)/[cos2(u) – sin2(u)]
= {2sin(u)cos(u)/cos2(u)} / {[cos2(u) – sin2(u)]/cos2(u)}
= {2sin(u)/cos(u)} / {1 - sin2(u)/cos2(u)}
= 2tan(u) / (1 - tan2(u))
Power Reducing Formulas
(16) sin2(u) = [1-cos(2u)]/2
We can easily arrive to this from (13.3)
(17) cos2(u) = [1+cos(2u)]/2
We can easily arrive to this from (13.2)
(18) tan2(u) = [1-cos(2u)]/[1+cos(2u)]
We can easily arrive to this by dividing (17) by (18)
Half Angle Formulas
(19) sin(u/2) = +/-[1-cos(u))/2]1/2
sin2(x) = [1-cos(2x)]/2 ------ Using (16)
=> sin2(u/2) = [1-cos(u)]/2 ------ let x = u/2
=> sin(u/2) = +/-[1-cos(u))/2]1/2
(20) cos(u/2) = +/-[1+cos(u))/2]1/2
Similar to (19), we can arrive this from (17)
(21.1) tan(u/2) = [1-cos(u)] / sin(u)
tan2(x) = [1-cos(2x)]/[1+cos(2x)] ------ Using (18)
=> tan2(u/2) = [1-cos(u)]/[1+cos(u)] ------ let x = u/2
=> tan2(u/2) = {[1-cos(u)]*[1-cos(u)]} / {[1+cos(u)]*[1-cos(u)]}
=> tan2(u/2) = [1-cos(u)]2 / [1-cos2(u)]
=> tan2(u/2) = [1-cos(u)]2 / sin2(u) ------ Using cos2(u) + sin2(u) = 1
=> tan(u/2) = [1-cos(u)] / sin(u)
(21.2) tan(u/2) = sin(u) / [1+cos(u)]
tan2(x) = [1-cos(2x)]/[1+cos(2x)] ------ Using (18)
=> tan2(u/2) = [1-cos(u)]/[1+cos(u)] ------ let x = u/2
=> tan2(u/2) = {[1-cos(u)]*[1+cos(u)]} / {[1+cos(u)]*[1+cos(u)]}
=> tan2(u/2) = [1-cos2(u)] / [1+cos(u)]2
=> tan2(u/2) = sin2(u) / [1+cos(u)]2 ------ Using cos2(u) + sin2(u) = 1
=> tan(u/2) = sin(u) / [1+cos(u)]
