Law of cosines

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Revision as of 15:48, 3 June 2011

Law of Cosines

Field: Geometry
Created By: rscott3 Website: [ ]

Law of Cosines

The law of cosines is a trigonometric extension of the Pythagorean Theorem.

1 Basic Description
- 1.1 Alternate Forms
2 A More Mathematical Explanation
- 2.1 Proof
  - 2.1.1 By the Pythagorean Theorem
  - 2.1.2 Using the Distance Formula
- 2.2 Example Problem
  - 2.2.1 Solution
3 Teaching Materials

Basic Description

The law of cosines is a formula that helps in solving triangles when two or three side lengths of a triangle are known. The formula combines the squares of two side lengths of a triangle and some offset, classified by the cosine of a particular angle, to calculate the square of the third side. For this reason, the law of cosines is often thought of as the generalization of the Pythagorean theorem, which only applies to right triangles. The law of cosines adds an extra term to the Pythagorean theorem so that a third side length of a triangle can be determined when there is no right angle.

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

The law of cosines is useful in solving triangles whenever at least two side lengths are known. It us also helpful when just three side lengths are known, and no angle measures. Given three elements of a triangle, there must be at least two side lengths given to successfully implement the law of cosines. If the triangle that needs to be solved has more than just one side length missing, the Law of sines is more useful.

Alternate Forms

Particularly when the law of cosines is being used to find a particular side length, that is when a certain angle measure is given, the law of cosines can be written in several different ways to help set up the equation to solve for the missing element of the triangle.

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

$b^{2} = a^{2} + c^{2} - 2ac \cos B$

$a^{2} = b^{2} + c^{2} - 2bc \cos A$

An alternate form of the law of cosines is particularly useful when solving a triangle when just the three side lengths are given. This for isolates the term with the cosine of the angle in it to make it easier to solve.

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

$\cos B = \frac{a^{2} + c^{2}- b^{2}}{2ac}$

$\cos A = \frac{b^{2} + c^{2}- a^{2}}{2bc}$

A More Mathematical Explanation

Proof

By the Pythagorean Theorem

An easy way to think of the law of cosines is as an extens [...]

Proof

By the Pythagorean Theorem

An easy way to think of the law of cosines is as an extension of the Pythagorean theorem for a right triangle:

$a^{2} + b^{2} = c^{2}$

By Pythagorean theorem, we know

$c^{2} = p^{2} + h^{2}$

But $p$ is just some portion of side length $a$ which is $r$ less than the length of $a$ . Substituting the difference gives us,

$c^{2} = (a-r)^{2} + h^{2}$

By Pythagorean theorem, we also know that

$h^{2} = b^{2} - r^{2}$

Substituting the appropriate values gives us,

$c^{2} = (a-r)^{2} + ( b^{2} - r^{2})$

Expanding the squared term gives us

$c^{2} = a^{2} -2ar + r^{2} + b^{2} - r^{2}$

Simplify for

$c^{2} = a^{2} -2ar + b^{2}$

And by the definition of cosine, we know that

$r=b \cos C$

Substituting this value in give us

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

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

Using the Distance Formula

Let $\vartriangle ABC$ be oriented so that $C$ is at the origin, and $B$ is at the point $(a,0)$ .

$\text{distance} = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$

$c$ is the distance from $A$ to $B$ .

Substituting the appropriate points into the distance formula gives us

$c = \sqrt {(a-b \cos C)^{2} + (0-b \sin C)^{2}}$

Squaring the inner terms, we have

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

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

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

Square both sides for

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

Example Problem

Complete the triangle using the law of cosines.

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

Solution

[show more][hide]

To find the side length $c$ ,

$c^{2} = 6^{2} + (6 \sqrt{2})^{2} -2 (6) (6 \sqrt{2}) \cos 45^\circ$

Simplify for

$c^{2} =36 + 36 (2) - 72 \sqrt{2}) \cos 45^\circ$

Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$ , substitution gives us

$c^{2} =36 + 36 (2) - 72 \sqrt{2} (\frac{1}{\sqrt{2}})$

Simplify for

$c^{2} =36 + 72 - 72$

$c^{2} =36$

Taking the square root of both sides gives us

$c =6$

Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure $B$ ,

$b^{2} = a^{2} + c^{2} - 2ab \cos B$

Substituting in the appropriate side lengths gives us

$(6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B$

Simplify for

$36 (2) = 36 + 36 - 72 \cos B$

$72 = 72 - 72 \cos B$

Subtracting $72$ from both sides gives us

$0 = - 72 \cos B$

Dividing both sides by $-72$ gives us

$0 = \cos B$

Using inverse trigonometry, we know that

$B = 90^\circ$

And we can find the last angle measure $A$ by subtracting the other two measures from $180^\circ$

$180^\circ - 90^\circ - 45^\circ = 45^\circ$

$A=45^\circ$

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Categories: Higher Education | Math Images In Progress | Geometry | Rscott3

@@ Line 7: / Line 7: @@
 ::<math> c^{2} = a^{2} + b^{2} - 2ab \cos C </math>
-When to use it: SAS, SSS.
+The law of cosines is useful in solving triangles whenever at least two side lengths are known. It us also helpful when just three side lengths are known, and no angle measures. Given three elements of a triangle, there must be at least two side lengths given to successfully implement the law of cosines. If the triangle that needs to be solved has more than just one side length missing, the [[Law of sines]] is more useful.
+===Alternate Forms===
+Particularly when the law of cosines is being used to find a particular side length, that is when a certain angle measure is given, the law of cosines can be written in several different ways to help set up the equation to solve for the missing element of the triangle.
+:*<math> c^{2} = a^{2} + b^{2} - 2ab \cos C </math>
+:*<math> b^{2} = a^{2} + c^{2} - 2ac \cos B </math>
+:*<math> a^{2} = b^{2} + c^{2} - 2bc \cos A </math>
+An alternate form of the law of cosines is particularly useful when solving a triangle when just the three side lengths are given. This for isolates the term with the cosine of the angle in it to make it easier to solve.
+:*<math> \cos C = \frac{a^{2} + b^{2}- c^{2}}{2ab}</math>
+:*<math> \cos B = \frac{a^{2} + c^{2}- b^{2}}{2ac}</math>
+:*<math> \cos A = \frac{b^{2} + c^{2}- a^{2}}{2bc}</math>
 |ImageDesc===Proof==
 ===By the Pythagorean Theorem===

Law of cosines

From Math Images

Revision as of 15:48, 3 June 2011

Contents

Basic Description

Alternate Forms

A More Mathematical Explanation

Proof

By the Pythagorean Theorem

Proof

By the Pythagorean Theorem

Using the Distance Formula

Example Problem

Solution

Teaching Materials

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