# Thread: Trigonometric Proof (Very Challenging)

1. ## Trigonometric Proof (Very Challenging)

Prove that cot(A) + cot(B) + cot(C) = (a^2 +b^2 +c^2) / 4(Area of ABC), where ABC is a traingle and a, b, c are opposite sides relative to the angles.

2. Originally Posted by strummer

Prove that cot(A) + cot(B) + cot(C) = (a^2 +b^2 +c^2) / 4(Area of ABC), where ABC is a traingle and a, b, c are opposite sides relative to the angles.
actually it's quite easy: let $\displaystyle S$ be the area of the triangle. then since $\displaystyle S=\frac{bc\sin A}{2}$ and: $\displaystyle bc\cos A = \frac{b^2 +c^2 - a^2}{2},$ we'll have: $\displaystyle \cot A= \frac{b^2 +c^2 - a^2}{4S}.$

similarly: $\displaystyle \cot B= \frac{a^2 + c^2 - b^2}{4S},$ and $\displaystyle \cot C= \frac{a^2 + b^2 - c^2}{4S}.$ now add these three relations and you're done!

3. Originally Posted by strummer
Prove that cot(A) + cot(B) + cot(C) = (a^2 +b^2 +c^2) / 4(Area of ABC), where ABC is a traingle and a, b, c are opposite sides relative to the angles.
hey mate,

I'll take on the challenge!

As I'm sure you know cot(x) = cos(x)/sin(x)

Thus for each A,B,C we need to resolve cot(x) or moreover sin(x), cos(x)

for cot(A)
(c^2 - (a^2 + b^2))/(2ac*sin(B))

Cosine Rule :
a^2 = b^2 + c^2 - 2bc*cos(A)
Thus cos(A) = (a^2 - (b^2 + c^2)) / (2bc)

Sine Rule :
a/sin(A) = b/sin(B)

Thus,
sin(A) =(a/b)sin(B)

Hence,

cot(A) = (((b^2 + c^2)-a^2) / 2bc)/ ((a/b)sin(B))
cot(A) = (( (b^2 + c^2)-a^2) / (2ac * sin(B)))

for cot(B)

Cosine Rule :
b^2 = a^2 + c^2 - 2ac*cos(B)
cos(B) = ((a^2 + c^2)-b^2) /(2ac)

Thus,

cot(B) = (((a^2 + c^2)-b^2) /(2ac))/sin(B)
cot(B) = (((b^2 + c^2)-b^2) / (2ac * sin(B)))

for cot(C)

Cosine Rule :
c^2 = a^2 + b^2 - 2ab*cos(C)
cos(C) = (c^2 - (a^2 + b^2))/(2ab)

Sine Rule:
b/sin(B) = c/sin(C)
Thus,
sin(C) = (c/b)sin(B)

Thus,

cot(C) = ((c^2 - (a^2 + b^2))/(2ab))/ ( (c/b)sin(B) )
cot(C) = (c^2 - (a^2 + b^2))/(2ac*sin(B))

Hence we now can define cot(A) + cot(B) + cot(C)

cot(A) + cot(B) + cot(C)
= (( (b^2 + c^2)-a^2) / (2ac * sin(B))) + cot(B) = (((a^2 + c^2)-b^2) / (2ac * sin(B))) + ((a^2 + b^2)-c^2)/(2ac*sin(B))
= (1/(2ac * sin(B))(a^2 + b^2 + c^2)

Area of ABC = (1/2)(ac * sin(B) --> ac*sin(B) = 2 *(Area of ABC)

Sub in,
cot(A) + cot(B) + cot(C) = (a^2 + b^2 + c^2)/ (2 * (2 * (Area of ABC)))
cot(A) + cot(B) + cot(C) = (a^2 + b^2 + c^2)/(4 x (Area of ABC))
= RHS

Thus true,

Hope this helps,

David

4. Originally Posted by David24
hey mate,

I'll take on the challenge!

As I'm sure you know cot(x) = cos(x)/sin(x)

Thus for each A,B,C we need to resolve cot(x) or moreover sin(x), cos(x)

for cot(A)
(c^2 - (a^2 + b^2))/(2ac*sin(B))

Cosine Rule :
a^2 = b^2 + c^2 - 2bc*cos(A)
Thus cos(A) = (a^2 - (b^2 + c^2)) / (2bc)

Sine Rule :
a/sin(A) = b/sin(B)

Thus,
sin(A) =(a/b)sin(B)

Hence,

cot(A) = (((b^2 + c^2)-a^2) / 2bc)/ ((a/b)sin(B))
cot(A) = (( (b^2 + c^2)-a^2) / (2ac * sin(B)))

for cot(B)

Cosine Rule :
b^2 = a^2 + c^2 - 2ac*cos(B)
cos(B) = ((a^2 + c^2)-b^2) /(2ac)

Thus,

cot(B) = (((a^2 + c^2)-b^2) /(2ac))/sin(B)
cot(B) = (((b^2 + c^2)-b^2) / (2ac * sin(B)))

for cot(C)

Cosine Rule :
c^2 = a^2 + b^2 - 2ab*cos(C)
cos(C) = (c^2 - (a^2 + b^2))/(2ab)

Sine Rule:
b/sin(B) = c/sin(C)
Thus,
sin(C) = (c/b)sin(B)

Thus,

cot(C) = ((c^2 - (a^2 + b^2))/(2ab))/ ( (c/b)sin(B) )
cot(C) = (c^2 - (a^2 + b^2))/(2ac*sin(B))

Hence we now can define cot(A) + cot(B) + cot(C)

cot(A) + cot(B) + cot(C)
= (( (b^2 + c^2)-a^2) / (2ac * sin(B))) + cot(B) = (((a^2 + c^2)-b^2) / (2ac * sin(B))) + ((a^2 + b^2)-c^2)/(2ac*sin(B))
= (1/(2ac * sin(B))(a^2 + b^2 + c^2)

Area of ABC = (1/2)(ac * sin(B) --> ac*sin(B) = 2 *(Area of ABC)

Sub in,
cot(A) + cot(B) + cot(C) = (a^2 + b^2 + c^2)/ (2 * (2 * (Area of ABC)))
cot(A) + cot(B) + cot(C) = (a^2 + b^2 + c^2)/(4 x (Area of ABC))
= RHS

Thus true,

Hope this helps,

David
I guess the OP now has a choice as to which path to follow .....