Trigonometry/Law of Cosines

< Trigonometry

Law of CosinesEdit


The Pythagorean theorem is a special case of the more general theorem relating the lengths of sides in any triangle, the law of cosines:[1]

a^2+b^2-2ab\cos{\theta}=c^2, \,

where θ is the angle between sides a and b.

Does the formula make sense?Edit

This formula had better agree with the Pythagorean Theorem when \displaystyle \theta=90^\circ.

So try it...

When \displaystyle \theta=90^\circ, \displaystyle \cos\theta=\cos 90^\circ=0

The \displaystyle -2ab\cos\theta=0 and the formula reduces to the usual Pythagorean theorem.


For any triangle with angles A, B, and C and corresponding opposite side lengths a, b, and c, the Law of Cosines states that

a^2=b^2 + c^2 - 2bc \cdot\cos A,
b^2=a^2 + c^2 - 2ac \cdot\cos B,
c^2=a^2 + b^2 - 2ab \cdot\cos C.



Dropping a perpendicular OC from vertex C to intersect AB (or AB extended) at O splits this triangle into two right-angled triangles AOC and BOC, with altitude h from side c.

First we will find the lengths of the other two sides of triangle AOC in terms of known quantities, using triangle BOC.

h=a sin B

Side c is split into two segments, with total length c.

OB has length a cos B
AO has length c - a cos B

Now we can use the Pythagorean Theorem to find b, since b2 = AO2 + h2.

\begin{matrix}b^2 & = & (c-a\cos B)^2+a^2\sin^2 B \\
\ &=& c^2-2ac\cos B +a^2\cos^2 B+a^2\sin^2 B\\
\ &=& a^2+c^2-2ac\cos B\end{matrix}

The corresponding expressions for a and c can be proved similarly.

The formula can be rearranged:

cos(C) = \frac{a^2+b^2-c^2}{2ab}

and similarly for cos(A) and cos(B).


This formula can be used to find the third side of a triangle if the other two sides and the angle between them are known. The rearranged formula can be used to find the angles of a triangle if all three sides are known. See Solving Triangles Given SAS.