Question

How do I prove that in a triangle with sides a, b c,

cot C - cot B = (b^2 - c^2) / (bcsinA)

Thanks.

Answer 1

Answer:

Assume triangle ABC with a opposite A, b opposite B, and c opposite C.

Also assume you have already proved the formula for the area of a triangle given 2 sides and an included angle. Let K = area of triangle ABC. Then by this formula the following are true:

1/2 ab sinC = K

1/2 ac sinB = K

1/2 bc sinA = K

That is given two sides and an included angle the area is 1/2 the product of the two sides times the sine of the included angle.

Using the Law of Cosines we also have:

a2 = b2+c2-2bccosA

b2 = a2+c2-2accosB

c2 = a2+b2-2abcosC

The trick is to add these three equations together, simplify, divide by 4, and then replace 1/2 ac with K/sinB (for example).

a2+b2+c2=2a2+2b2+2c2-2bccosA-2accosB-2abcosC

This can be rewritten as:

a2+b2+c2=2bccosA + 2accosB + 2abcosC

Divide every term by 4:

I. (a2+b2+c2)/4 = 1/2 bc cosA + 1/2 ac cosB + 1/2 ab cosC

But 1/2 bc = K/sinA; 1/2 ac = K/sinB ; 1/2 ab = K/sinC (these from the earlier mentioned formulas for area)

Substitute into I giving on the right side: (K/sinA) cosA + (K/sinB) cosB + (K/sinC) cosC

Replace cos X / sin X with cot X and on right side we have:

= KcotA + KcotB + KcotC

Now divide by K (which is the area of the triangle)

II. (a2+b2+c2)/4K = cotA + cotB + cotC

Hence, Proved