# Pythagoras's theorem as a special case of the law of cosines

I heard that the Pythagorean theorem is a special case of the more general theorem relating the lengths of sides in any triangle, the law of cosines?

• What is your Question? Mar 8 '14 at 11:08
• Yes. Pythagoras Theorem can be seen as a very particular case of the cosines law, though historically it is not so. Mar 8 '14 at 11:16
• @fcpatidar11 welcome to math.stackexchange! You might want to have a look at the short intro to the site mechanics, to get the hang of how things work around here :) Mar 8 '14 at 13:17

## 2 Answers

The law of cosines is: $$c^2 = a^2 + b^2 \;-\; 2\!\cdot\!a\!\cdot\!b\!\cdot\!\cos\theta$$ where $\theta$ is the angle between the sides $a$ and $b$.
Now, when this angle is a right angle ($90^\circ$, or $\frac{\pi}{2}$), its cosine is $0$, so the entire last term is multiplied by $0$ and vanishes, leaving only the usual Pythagorean Theorem for the right triangle with legs $a$ and $b$ and the hypotenuse $c$: $$c^2 = a^2 + b^2$$ Simple as that.

Yup: it states that $$a=\sqrt{b^2+c^2-2bc\cos\alpha}$$ in standard trigonometric notation (where $a,b,c$ are the sides of the triangle, and $\alpha$ is the angle which is opposite to $a$).

A simple proof can be given with vector calculus: being $\vec{a}=\vec{b}-\vec{c}$, squaring this relation you obtain: $$\vec{a}\cdot\vec{a}\equiv a^2 = b^2 + c^2 - 2 \vec{b}\cdot \vec{c}$$ which is the formula above substituting the definition of scalar product!

• how do you prove $a.b=|a||b|\cos\theta$?.You need law of cosine to do that!.This is circular reasoning May 14 '18 at 21:11
• @Cloud JR actually that is my definition of scalar product May 14 '18 at 22:20