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Trigonometry Completion of the proof of the Cosine Rule

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Hello my friends,

I posted this picture as a proof of the Cosine Rule in another thread,



however after having a closer look at it, I believe it is incomplete. It works by drawing a segment from one of the vertices so that this segment is perpendicular to one side of the triangle, and then applying Pythagoras' Theorem.

However, if you have an obtuse-angled triangle, it is impossible to draw a segment from one of the acute vertices to make a right-angle triangle. So how is it possible to prove this relationship:

\(\displaystyle \displaystyle c^2 = a^2 + b^2 - 2\,a\,b\cos{(C)}\)

when C is an obtuse angle?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Please refer to the following diagram:

proveit.jpg

Note: I have used \(\displaystyle \sin(\pi-\theta)=\sin(\theta)\) and \(\displaystyle \cos(\pi-\theta)=-\cos(\theta)\).

By Pythagoras, we now have:

\(\displaystyle (a-b\cos(\theta))^2+(b\sin(\theta))^2=c^2\)

\(\displaystyle c^2=a^2-2ab\cos(\theta)+b^2\cos^2(\theta)+b^2\sin^2(\theta)\)

\(\displaystyle c^2=a^2+b^2-2ab\cos(\theta)\)