Trigonometry/Law of Cosines

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Law of CosinesEdit

Law-of-cosines1.svg

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 \theta is the angle between sides a and b .

Does the formula make sense?Edit

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

So try it...

When \theta=90^\circ , \cos(\theta)=\cos(90^\circ)=0

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

PermutationsEdit

For any triangle with angles A,B,C and corresponding opposite side lengths a,b,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)

ProofEdit

Law-of-cosines2.svg

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 \overline{AB}\cos(B)
AO has length c-a\cos(B)

Now we can use the Pythagorean Theorem to find b , since b^2=\overline{AO}^2+h^2 .

b^2 =\bigl(c-a\cos(B)\bigr)^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)

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) .

ApplicationsEdit

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.

NotesEdit