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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?

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    $\begingroup$ What is your Question? $\endgroup$ Mar 8 '14 at 11:08
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    $\begingroup$ Yes. Pythagoras Theorem can be seen as a very particular case of the cosines law, though historically it is not so. $\endgroup$
    – DonAntonio
    Mar 8 '14 at 11:16
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    $\begingroup$ @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 :) $\endgroup$
    – Brightsun
    Mar 8 '14 at 13:17
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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.

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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!

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

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