sin(a+b) and cos(a+b)

In today's blog, I plan to go over a basic property of sine and cosine that I will use later to find the derivatives of sine and cosine.

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