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The Law of Cosines $$a^2=b^2+c^2-2bc\cos\theta$$ looks like an extension of the Pythagorean theorem, but what does the second part ($-2bc\cos\theta$) actually do?

Blue
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MathBun
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    It takes into account the fact that the angle is not always $90°$ (as in Pythagoras' theorem) :-) – Dominique Mar 21 '23 at 08:12
  • Draw (for example) an acute triangle with a horizontal base. Let the left and right vertices be denoted $~b~$ and $~c~$ respectively, and left the vertex at the top of the triangle be denoted $~a.~$ Let $~A, ~B, ~C~$ denote the sides opposite vertices (i.e. angles) $~a, ~b, ~c~$ respectively. Construct the altitude from $~b~$ to $~\overline{AC}.~$ Then $$\left[ ~C\sin(a) ~\right]^2 + \left[ ~B - C\cos(a) ~\right]^2 = A^2,$$ by the pythagorean theorem. The law of cosines is an immediate consequence of this. – user2661923 Mar 21 '23 at 08:17
  • @Dominique your answer is at a much higher level of abstraction than I wanted to get – MathBun Mar 21 '23 at 08:36
  • @user2661923 thanks! You helped a lot – MathBun Mar 21 '23 at 09:09
  • Related (duplicate?): "Where does the $−2$ term come from in the cosine law?". See, in particular, my answer. – Blue Mar 21 '23 at 09:13
  • In vector terms it's the cross-term from squaring viz. $\vec{a}=\vec{b}-\vec{c}\implies a^2=b^2+c^2-2\vec{b}\cdot\vec{c}$. – J.G. Mar 21 '23 at 09:21

2 Answers2

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The quantity $2bc⋅\cos(\theta)$ can be considered as a correction factor which covers all the possible cases, for example:

  • $\theta = 90° \implies \cos(\theta)=0$ which is indeed the Pythagorean theorem $a^2=b^2+c^2$

  • $\theta = 0° \implies \cos(\theta)=1$ which is $a^2=b^2+c^2-2bc =(b-c)^2$ and indeed $|a|=|b-c|$

  • $\theta = 180° \implies \cos(\theta)=-1$ which is $a^2=b^2+c^2+2bc =(b+c)^2$ and indeed $a=b+c$

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user
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It can be viewed as an extension of the Pythagorean Theorem and one way to directly prove it involves that most famous of Theorems.

What the second part "does" is "adjust" for the triangle not being a right triangle.

Let's assume you have the acute triangle $\triangle ABC$ with $AD$ being the height.

Then by Pythagoras, $b^2=AD^2+CD^2$.

But $CD=a-BD$ and $AD^2=c^2-BD^2$.

So $b^2=c^2-BD^2+(a-BD)^2=a^2+c^2-2a*BD$.

But $\cos B=\Large\frac{c}{BD}$ $\Rightarrow BD=c*\cos B$ and if we substitute above we obtain $b^2=a^2+c^2=2ac\cos B$ as desired. Taking the other two heights, we obtain the law for the other angles.

Similarly we can work on an obtuse triangle.

MathematicianByMistake
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