## Mathematical Proof of Cosine Identity (cosine) Theorem

Last summer I was randomly surfing on internet at home and saw cosine theory on wiki and wanted to proof the identities by myself. However, despite having a good trigonimetry and geometry background, I couldn’t remember the proof. Honestly I got a bit sad as I realized that it is being more than ten years since high school. I searched for its proof and I found a nice geometric proof. This time I got angry as I wasn’t able to find this geometric proof. It was time to get my hands dirty.

Here is a perpendicular triangle.

Step-1

Let’s say A is divided by a line segment

|AD| = t,

then |AB| = t*cos(A), BD = t*sin(A)

Now let’s draw a hight from D to |AC| named H so we have similarity between the big triangle and the small one; (ABC) ~ (DHC)

|HD| = t * sin(B),

|DC| = t * sin(B) / cos(A+B),

|HC| = t * sin(B) * sin(A+B) / cos(A+B)

|HA| = t * cos(B)

Step-2

Looking at the big triangle (ABC), its hypotenus is equal to |AH| + |HC| which also is also equal to |AB| / cos(A+B)

Step-3

Multiply both sides with cos(A+B), get rid of t and organize the equation

step-4, 5, 6, 7, 8

Organizing the equation

step-9

Here we come the the most interesting part, which we have a parabolic equation which means we are close to the solution

Steps 10,11

are the discriminant solution for the parabolic equation

Step 12 is the simplification of inside the root

And finaly step 13,

As Lois warned, as sin(A)*sin(B) comes out of the root it has one minus root

cos(A+B) = cos(A)*cos(B) – sin(A)*sin(B)

## 1 COMMENT

Hevisin(A)*sin(B) comes out from the root so it is actually ±sin(A)*sin(B). thanks for warning Lois!