## Definition

Both Sine rule and Cosine rule relate the sides of a triangle with its angles. Cosine rule (or Law of Cosines) is

\[ c^2 = a^2 + b^2 - 2ab \cos \gamma,\]

which can also be rewritten as

\[ \cos \gamma = \frac {a^2 + b^2 - c^2}{2ab}.\]

## Technique

## In triangle \( ABC\), we have \( \angle BAC = \frac {\pi}{4}\), \( BC = \sqrt{5}\) and \( AB = 3\). Determine \( AC\).

Applying the cosine rule on \( \angle BAC\), we get that \( AB^2 + AC^2 - 2 AB \cdot AC \cdot \cos \frac {\pi}{4} = BC^2\), which gives \( 0=AC^2 - 3 \sqrt{2} AC + 4 = (AC-\sqrt{2})(AC-2\sqrt{2})\). Hence, \( AC=\sqrt{2} \text{ or } 2\sqrt{2}\). \( _\square \)

*Note: *This is similar to the 'ambiguous case' of Sine rule, since we have \( 3 \sin \frac {\pi}{4} < \sqrt{5} < 3\), which is the condition \( x \sin \alpha < a < c\).

## Suppose \( a, b\) and \( c\) are positive reals such that \( a^2 = b^2+c^2 - bc\), \( b^2 = c^2 + a^2 - ac\) and \( c^2 = a^2 + b^2 - ab\). Show that \( a = b = c\).

Solution: Since \( c^2 = a^2 + b^2 - ab < a^2 + b^2 + 2ab = (a+b)^2\), it follows that \( c < a+b\) and we have similar inequalities for other variables. Hence, the numbers \( a, b\) and \( c\) satisfy the triangle inequality, and there exists a triangle \( ABC\) such that \( AB=c, BC=a, CA=b\). As such \( \cos \angle ABC = \frac {a^2 + c^2 - b^2}{2ac} = \frac {a^2 + c^2 - (a^2+c^2-ac)}{2ac} = \frac {1}{2}\), which gives us that \( \angle ABC = 60^\circ\). Similarily, we have \( \angle BCA = 60^\circ, \angle CAB = 60^\circ\), which show that triangle \( ABC\) is equilaterial, thus \( a=b=c\). \( _\square \)

Note:This may also be done directly by summing up the 3 equations to get that \( (a-b)^2 + (b-c)^2 + (c-a)^2 =0\).

## Application and Extensions

## Assuming the Cosine Rule, prove the Pythagorean theorem.

In right triangle \( ABC\) with \( \angle ACB = 90^\circ\), we have \( \cos \angle ABC = \frac {a^2+b^2-c^2}{2ab}\), hence \( a^2+b^2-c^2=0\), which gives that \( a^2 + b^2 = c^2\). \( _\square \)

## [Apollonius' Theorem] In triangle \( ABC\), \( D\) is the midpoint of \( BC\). Show that \( AB^2 + AC^2 = 2(AD^2+BD^2)\).

Solution: Apply the cosine rule to triangle \( ABD\), we get \( AB^2 = AD^2 + BD^2 - 2 AD \cdot BD \cos \angle ADB\). Applying the cosine rule to triangle \( ADC\), we get \( AC^2 = AD^2 + CD^2 - 2 AD \cdot DC \cos \angle ADC\). Since \( \angle ADB + \angle ADC = 180^ \circ\), so \( \cos \angle ADB = - \cos \angle ADC\). Adding the first 2 equations, and using the fact that \( BD=CD\), we get that \( AB^2 + AC^2 = AD^2 + BD^2 + AD^2 + CD^2 = 2(AD^2 + BD^2)\). \( _\square \)

## (IMO'68) Determine all triangles whose side lengths are consecutive positive integers, and one of the angles is twice of the other.

Solution: In triangle \( ABC\), let \( BC=n, CA=n+1, AB = n+2\). Let \( \alpha, \beta\) and \( \gamma\) be the angles at vertices \( A, B\) and \( C\), respectively. Then \( \gamma > \beta > \alpha\). Applying the cosine rule, we obtain \( \cos \alpha = \frac {(n+1)^2 + (n+2)^2 - n^2}{2(n+1)(n+2)} = \frac {n^2+6n+5}{2(n+1)(n+2)} = \frac {n+5}{2n+4}\). In a similar manner, we can obtain \( \cos \beta = \frac {n+1}{2n}\) and \( \cos \gamma = \frac {n-3}{2n}\). The double angle formula states that \( \cos 2 \theta = 2 \cos ^2 \theta -1\). Consider the following cases:

Case 1.\( \gamma = 2 \alpha\). Then, \( \frac {n-3}{2n} = 2 (\frac {n+5}{2n+4})^2 - 1\). Clearing denominators and expanding, we get \( 0 = 2n^3 - n^2 - 25n - 12 = (n-4) (2n+1)(n+3)\). The only positive integer solution is \( n=4\).

Case 2.\( \gamma = 2 \beta\). Then, \( \frac {n-3}{2n} = 2 (\frac {n+1}{2n})^2 - 1\). Clearing denominators and expanding, we get \( 0 = 2n^3 + n^2 - n = n(2n -1)(n+1)\). This has no positive integer solutions.

Case 3.\( \beta = 2 \alpha\). Then, \( \frac {n+1}{2n} = 2 (\frac {n+5}{2n+4})^2 - 1\). Clearing denominators and expanding, we get \( 0=2n^3 +3n^2-9n+4 = (n-1) (2n^2+5n-4)\). The only positive integer solution is \( n=1\). However, checking this triangle, we get the degenerate \( 1-2-3\) triangle, hence we ignore this solution.Thus, the only triangle is the \( 4-5-6\) triangle. \( _\square \)

## Comments

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TopNewestits invalid to prove pythagorean theorem using Law of Cosines. it uses the fact that

\( \sin^2 \theta + \cos^2 \theta = 1 \)

which is in fact a result of pythagorean theorem. – Aneesh Kundu · 2 years, 4 months ago

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THANKS..... – Klaurence Patrick Cruz · 1 year, 7 months ago

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Really interesting! – Jordi Bosch · 2 years, 5 months ago

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Thanks for sharing! It is really helpful.

However, I would like to know how to make those cool boxes?

Thanks,

Daniel – Daniel Liu · 2 years, 7 months ago

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– حكيم الفيلسوف الضائع · 2 years, 5 months ago

The \(\LaTeX\) code is: \square, and there's also a variant \blacksquare which produces: \(\quad \square\quad\blacksquare.\)Log in to reply