Law of cosines

(Difference between revisions)
 Revision as of 15:41, 3 June 2011 (edit)← Previous diff Revision as of 14:43, 20 June 2011 (edit) (undo)Next diff → Line 3: Line 3: |Image=Law_of_cosines_pictiure.jpg |Image=Law_of_cosines_pictiure.jpg |ImageIntro=The law of cosines is a trigonometric extension of the Pythagorean Theorem. |ImageIntro=The law of cosines is a trigonometric extension of the Pythagorean Theorem. - |ImageDescElem=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. + |ImageDescElem=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 and a third term involving 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. - Given a triangle with side lengths, $a,b,c$ and angle measures $A,B,C$, + Given a triangle with side lengths, $a,b,c$ and angle measures $A,B,C$, the law of cosines states ::$c^{2} = a^{2} + b^{2} - 2ab \cos C$ ::$c^{2} = a^{2} + b^{2} - 2ab \cos C$ - 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. + The law of cosines is useful in solving triangles whenever at least two side lengths are known. If the triangle that needs to be solved has less than just two known lengths, the [[Law of sines]] is more useful. Line 25: Line 25: :*$a^{2} = b^{2} + c^{2} - 2bc \cos A$ :*$a^{2} = b^{2} + c^{2} - 2bc \cos A$ - 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. + 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 for the angle. :*$\cos C = \frac{a^{2} + b^{2}- c^{2}}{2ab}$ :*$\cos C = \frac{a^{2} + b^{2}- c^{2}}{2ab}$ Line 105: Line 105: ==Example Problem== ==Example Problem== - Complete the triangle using the law of cosines. + Solve the triangle using the law of cosines. Line 121: Line 121: Simplify for Simplify for - ::$c^{2} =36 + 36 (2) - 72 \sqrt{2}) \cos 45^\circ$ + ::$c^{2} =36 + 36 (2) - 72 \sqrt{2} \cos 45^\circ$ Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$, substitution gives us Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$, substitution gives us Line 130: Line 130: ::$c^{2} =36 + 72 - 72$ ::$c^{2} =36 + 72 - 72$ - ::$c^{2} =36$ ::$c^{2} =36$ Line 140: Line 139: ---- ---- - Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure $B$, + To solve for angle $B$, we can orient the triangle differently and use the law of cosines in the form [[Image:SAS_solution.jpg|right]] [[Image:SAS_solution.jpg|right]] Line 149: Line 148: ::$(6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B$ ::$(6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B$ - - Simplify for ::$36 (2) = 36 + 36 - 72 \cos B$ ::$36 (2) = 36 + 36 - 72 \cos B$

Revision as of 14:43, 20 June 2011

Law of Cosines

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

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 and a third term involving 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.

Given a triangle with side lengths, $a,b,c$ and angle measures $A,B,C$, the law of cosines states

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

The law of cosines is useful in solving triangles whenever at least two side lengths are known. If the triangle that needs to be solved has less than just two known lengths, 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 for the angle.

• $\cos C = \frac{a^{2} + b^{2}- c^{2}}{2ab}$
• $\cos B = \frac{a^{2} + c^{2}- b^{2}}{2ac}$
• $\cos A = \frac{b^{2} + c^{2}- a^{2}}{2bc}$

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

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

or

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

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

Solve the triangle using the law of cosines.

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

Solution

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

$c =6$

To solve for angle $B$, we can orient the triangle differently and use the law of cosines in the form

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

Substituting in the appropriate side lengths gives us

$(6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B$
$36 (2) = 36 + 36 - 72 \cos B$
$72 = 72 - 72 \cos B$

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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