Law Of Cosines

In trigonometry, the law of cosines 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 sides of the triangle and A, B, and C the 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 angles 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 \; iff \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.

Derivation (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 b − a cos C. This makes two right triangles, one with legs a sin C and b − a 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 - 2 ab \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\,
because
\sin^2 C + \cos^2 C=1.\,

Law of cosines using vectors

Vector triangle
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} \;
since
\mathbf{(b - c)\cdot (b - c) = b\cdot b - 2 b\cdot c + c\cdot c}. \;
Using the dot product, we simplify this into
\mathbf{|a|^2 = |b|^2 + |c|^2 - 2 |b||c|}\cos \theta. \;

See also

External link

 

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