This is part of the larger story of using Taylor Series to come up with an equation for sine and cosine that does not depend on Euclid.
Lemma 1: Area of a triangle = (1/2)(ab)(sin C)
Proof:
Case I: C is less than 90 degrees.
(1) Let AD be perpendicular to BC
(2) We know that the area of triangle ABC is (1/2)ah (where a = the length of BC) [See here for details if needed]
(3) Now, we also know that:
sin C = h/b [Definition of sine, see here]
So that:
h = b sin C
(4) This gives us:
Area of triangle ABC = (1/2)ab*sin C [By substituting (#3) for (#2)]
Case II: C = 90 degrees
(1) Sine 90 degrees = 1 (see here for details if needed)
(2) In this case, b = h.
(3) So, the area is (1/2)ah = (1/2)ab = (1/2)ab sin 90 degrees = (1/2)ab sin C
Case III: C is greater than 90 degrees
(1) Let AD be perpendicular to BC so that B,C,D are colinear.
(2) We know that the area of triangle ABC is (1/2)ah (where a = the length of BC) [See here for details if needed]
(3) Now, we also know that:
sin C = h/b [Definition of sine, see here; since sin C = sin (180 - C).
So that:
h = b sin C
(4) This gives us:
Area of triangle ABC = (1/2)ab*sin C [By substituting (#3) for (#2)]
QED
Theorem 1: sin(a+b) = cosBsinA + cosAsinB
Proof:
(1) Let QN be a line that is perpendicular with PR.
(2) The area of triangle PQR = area of triangle PQN + area of triangle RQN
(3) By Lemma 1 above, we have:
Area of triangle PQR = (1/2)rp*sin(A+B)
Area of triangle PQN = (1/2)rh*sin(A)
Area of triangle RQN = (1/2)ph*sin(B)
(4) Combing step #2 with step #3 gives us:
(1/2)rp*sin(A+B) = (1/2)rh*sin(A) + (1/2)ph*sin(B)
(5) Multiplying 2 to both sides gives us:
rp*sin(A+B) = rh*sin(A) + ph*sin(B)
(6) Dividing both sides by rp gives us:
sin(A+B) = (h/p)*sin(A) + (h/r)*sin(B)
(7) Since QN is perpendicular to PR, we also know that (see here for definition of cosine):
cos B = h/p
cos A = h/r
(8) So that we have:
sin(A+B) = cos(B)*sin(A) + cos(A)sin(B)
QED
Lemma 2: Law of Cosines
c2 = a2 + b2 - 2ab*cos c
Proof:
Case I: Obtuse Angle (a triangle with one angle greater than 90 degrees)
(1) Let C be the obtuse angle (by C, I mean ∠ ACB)
(2) Let d = a + e
(3) Using Pythagorean Theorem (see here):
c2 = d2 + h2
b2 = e2 + h2
(4) So, we have:
c2 = (a+e)2 + h2 = a2 + 2ae + e2 + h2 =
= a2 + 2ae + b2
(5) Now, cos C = -cos(180-C) = -e/b [See here for details if needed]
So, we also have that:
e = -b*(cos C)
(6) Combining step #5 with step #6 gives us:
c2 = a2 + b2 - 2ab*(cos C).
Case II: Right triangle
In this case, the Pythagorean Theorem applies and we have:
c2 = a2 + b2
Now cos 90 degrees = 0 (see here if needed), so we also have:
c2 = a2 + b2 - 2ab*(cos C)
Case III: Acute triangle
(1) Let c be the acute angle.
(2) We will need to consider two cases. One where B is an acute angle and one where B is an obtuse angle.
(3) Let h be the height of the triangle.
(4) Let e represent the segment CD.
(5) Using Pythagorean Theorem, we have:
c2 = d2 + h2
b2 = e2 + h2
(6) Let a represent the segment BC.
(7) We see that in both cases,
d2 = (e-a)2 since:
(a) For the first triangle, a = d + e which implies -d = e - a.
(b) For the second triangle, e = d + a which implies d = e - a.
(8) And from this, we have:
d2 = e2 - 2ae + a2
which gives us:
d2 - e2 = a2 - 2ae
(9) From step #5, we have:
h2 = b2 - e2
So we also have:
c2 = d2 + h2 = d2 + b2 - e2 =
= a2 - 2ae + b2
(10) Now, cos C = e/b [See here for definition of cosine]
So we have: e = b cos C
And this then gives us:
c2 = a2 + b2 - 2ab(cos C)
QED
Theorem 2: cos(A+B) = cosA*cosB - sinA*sinB
NOTE: I will use the same diagram as I used for Theorem 1
Proof:
(1) (x+y)2 = r2 + p2 - 2rpcos(A+B) [See Law of Cosines above]
(2) If we substract (x+y)2 from both sides and add 2rp*cos(A+B) to both sides, we get:
2rp*cos(A+B) = r2 + p2 - (x+y)2 =
= r2 + p2 -x2 - 2xy - y2 =
= (r2 - x2) + (p2 - y2) - 2xy
(3) From the Pythagorean Theorem (see here), we get:
r2 = h2 + x2
and
p2 = h2 + y2
so that:
h2 = r2 - x2
and
h2 = p2 - y2
(4) Combining the results of step #3 with step #2 gives us:
2rp*cos(A+B) = r2 + p2 - (x+y)2 =(h2 + x2) + (h2 + y2) + 2xy - x2 - y2 = 2h2 - 2xy
(5) Dividing both sides by 2*rp gives us;
cos(A+B) = h2/(rp) - (xy)/(rp)
(6) Now from the definitions of sine and cosine (see here), we have:
cos A = h/r
cos B = h/p
sin A = x/r
sin B = y/p
(7) Combinining step #6 with step #5 gives us:
cos (A+B) = cosA*cosB - sinA*sinB
QED
References
- David Joyce, Area of a Triangle
- David Joyce, Law of Cosines
- Davis Associates, Pythagorean Theorem
2 comments :
On the proof for law of cosine, at one point it says cos0degrees=0 where in fact it should say cos90degrees=0
Hi Shawn,
Thanks for noticing that. I just fixed the typo.
-Larry
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