Law of cosines

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In trigonometry, the law of cosines (also known as the cosine formula) is a statement about arbitrary triangles which generalizes the Pythagorean theorem by correcting it with a term proportional to the cosine of the opposing angle. Let a, b, and c be the lengths of the sides of the triangle and let A, B, and C the respective angles opposite those sides. Then,

c^{2}=a^{2}+b^{2}-2ab\cos C.\;

This formula is useful for computing the third side of a triangle when two sides and their enclosed angle are known, and in computing the angles of a triangle if all three sides are known.

The law of cosines also shows that

c^{2}=a^{2}+b^{2}\;   if and only if   \cos C=0.\;

The statement cos C = 0 implies that C is a right angle, since a and b are positive. In other words, this is the Pythagorean theorem and its converse. Although the law of cosines is a broader statement of the Pythagorean theorem, it isn't a proof of the Pythagorean theorem, because the law of cosines derivation given below depends on the Pythagorean theorem.


Using vectors and vector dot products, we can easily prove the law of cosines. If we have a triangle with vertices A, B, and C whose sides are the vectors a, b, and c, we know that:

{\mathbf  {a=b-c}}\;


{\mathbf  {(b-c)\cdot (b-c)=b\cdot b-2b\cdot c+c\cdot c}}.\;

Using the dot product, we simplify the above into

{\mathbf  {|a|^{2}=|b|^{2}+|c|^{2}-2|b||c|}}\cos \theta .\;

Alternative proof (for acute angles)

Let a, b, and c be the sides of the triangle and A, B, and C the angles opposite those sides. Draw a line from angle B that makes a right angle with the opposite side b. If the length of that line is x, then sin C = x/a, which implies x = a sin C.

That is, the length of this line is a sin C. Similarly, the length of the part of b that connects the foot point of the new line and angle C is a cos C. The remaining length of b is ba cos C. This makes two right triangles, one with legs a sin C and ba cos C and hypotenuse c. Therefore, according to the Pythagorean theorem:

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


\sin ^{2}C+\cos ^{2}C=1.\,

Finding the angles when the sides are known

By transposing the identity

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

we can find the angle C when the three sides a, b, and c are known:

C=\arccos {\frac  {a^{2}+b^{2}-c^{2}}{2ab}}.

Isosceles case

When a=b, i.e., when the triangle is isosceles with the two sides incident to the angle C equal, the law of cosines simplies significantly. Namely, because a^{2}+b^{2}=2a^{2}=2ab, the law of cosines becomes

\cos C=1-{\frac  {c^{2}}{2a^{2}}}.\;

See also

External link

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