Lesson PROOF OF THE LAW OF COSINES

Source code of 'PROOF OF THE LAW OF COSINES'

This Lesson (PROOF OF THE LAW OF COSINES) was created by by Theo(13342)

This lesson provides an overview of the Proof of the Law Of Cosines.

REFERENCES

<a href = "http://www.clarku.edu/~djoyce/trig/laws.html" target = "_blank">http://www.clarku.edu/~djoyce/trig/laws.html</a>
<a href = "http://mathproofs.blogspot.com/2006/06/law-of-cosines.html" target = "_blank">http://mathproofs.blogspot.com/2006/06/law-of-cosines.html</a>
<a href = "http://www.mathamazement.com/Lessons/Pre-Calculus/06_Additional-Topics-in-Trigonometry/law-of-cosines.html" target = "_blank">http://www.mathamazement.com/Lessons/Pre-Calculus/06_Additional-Topics-in-Trigonometry/law-of-cosines.html</a>
<a href = "http://www.themathpage.com/aTrig/law-of-cosines.htm" target = "_blank">http://www.themathpage.com/aTrig/law-of-cosines.htm</a>
<a href = "http://pages.pacificcoast.net/~cazelais/173/law-sines-cosines.pdf" target = "_blank">http://pages.pacificcoast.net/~cazelais/173/law-sines-cosines.pdf</a>
<a href = "http://pballew.net/lawofcos.htm" target = "_blank">http://pballew.net/lawofcos.htm</a>
<a href = "http://www.kkuniyuk.com/M1410602.pdf" target = "_blank">http://www.kkuniyuk.com/M1410602.pdf</a>

There are 3 cases to the proof.

They are:

CASE 1

All angles in the triangle are acute.

CASE 2

There is one right angle in the triangle.

CASE 3

There is one obtuse angle in the triangle.

LAW OF COSINES EQUATIONS

They are:

{{{c^2 = a^2 + b^2 - 2*a*b*cos(C)}}}

{{{a^2 = b^2 + c^2 - 2*b*c*cos(A)}}}

{{{b^2 = a^2 + c^2 - 2*a*c*cos(B)}}}

The proof will be for:

{{{c^2 = a^2 + b^2 - 2*a*b*cos(C)}}}

This is based on the assumption that, if we can prove that equation, we can prove the other equations as well because the only difference is in the labeling of the points on the same triangle.

PROOF OF LAW OF COSINES EQUATION

CASE 1

All angles in the triangle are acute.

A picture of our triangle is shown below:

Our triangle is triangle ABC.

We drop a perpendicular from point B to intersect with side AC at point D.

That creates 2 right triangles (ABD and CBD).

Side b from triangle ABC is equal to side d from triangle ABD plus side e from triangle CBD.

Since Triangle ABD and CBD are both right triangles, we can use the Pythagorean Formula to identify the relationship between the hypotenuse and each leg of each triangle.

For Triangle ABD, we get:

{{{c^2 = d^2 + h^2}}} *****

***** means this is the equation that will eventually be equal to {{{c^2 = a^2 + b^2 - 2*a*b*cos(C)}}}

Since b = d + e, this means that d = b - e.

Substituting in our equation, we get:

{{{c^2 = (b-e)^2 + h^2}}} *****

Since we also know that sin(C) in Triangle CBD = {{{h/a}}}, we can solve for h to get:

h = a*sin(C).

We can substitute for h in our equation to get:

{{{c^2 = (b-e)^2 + (a*sin(C))^2}}} *****

Since {{{(a*sin(C))^2}}} is the same as {{{a^2 * sin^2(C)}}}, then our equation becomes:

{{{c^2 = (b-e)^2 + a^2 * sin^2(C)}}} *****

If we complete the squaring of (b-e), then it becomes {{{(b-e)^2 = b^2 - 2*b*e + e^2}}}.

If we plug this into our equation, then we get:

{{{c^2 = b^2 - 2*b*e + e^2 + a^2 * sin^2(C)}}} *****

Since we know that cos(C) in Triangle CBD is equal to {{{e/a}}}, we can solve for e to get e = a * cos(C)

If we substitute for e in our equation, then we get:

{{{c^2 = b^2 - 2*b*a*cos(C) + (a*cos(C))^2 + a^2 * sin^2 (C)}}} *****

Since {{{(a*cos(C))^2}}} is the same as {{{a^2 * cos^2(C)}}}, then we can substitute in our equation to get:

{{{c^2 = b^2 - 2*b*a*cos(C) + (a^2 * cos^2(C)) + (a^2 * sin^2(C))}}} *****

Since we know that b*a is the same as a*b, then we can rearrange terms in our equation to get:

{{{c^2 = b^2 - 2*a*b*cos(C) + (a^2 * cos^2(C)) + (a^2 * sin^2(C))}}} *****

We can factor out the {{{a^2}}} in {{{(a^2 * cos^2(C)) + (a^2 * sin^2(C))}}} to get {{{a^2 * (cos^2(C) + sin^2(C))}}}

Since we know that cos2(C) + sin2(C) = 1, then we get {{{(a^2 * (cos^2(C)) + sin^2(C))}}} = {{{a^2}}}

If we substitute {{{a^2}}} for {{{(a^2 * cos^2(C))}}} + {{{a^2 * sin^2(C)}}} in our equation, then we get:

{{{c^2 = b^2 - 2*a*b*cos(C) + a^2}}} *****

By rearranging terms, this becomes the same as:

{{{c^2 = a^2 + b^2 - 2*a*b*cos(C)}}} *****

Since this is exactly the same as the original equation we were trying to prove, then we are done.

This concludes the proof for Case 1.

CASE 2

There is one right angle in the triangle.

This makes it a right triangle.

In a right triangle, the Pythagorean Formula prevails.

We start off with our original equation of {{{c^2 = a^2 + b^2 - 2*a*b*cos(C)}}}.

Since c is the hypotenuse of this right triangle, then angle C is opposite side c and is equal to 90 degrees.

If angle C is equal to 90 degrees, then cos(C) = cos(90) = 0.

This makes 2*a*b*cos(C) equal to 0.

The equation {{{c^2 = a^2 + b^2 - 2*a*b*cos(C)}}} becomes {{{c^2 = a^2 + b^2}}} which is exactly the Pythagorean formula.

The picture of our right triangle is shown below:

Since the Pythagorean formula prevails in a right triangle, and the Pythagorean Formula is a special case of our original equation, then we are done.

{{{c^2 = a^2 + b^2 - 2*a*b*cos(C)}}} becomes the same as {{{c^2 = a^2 + b^2}}} when cos(C) = 0.

This concludes the proof for case 2.

CASE 3

There is one obtuse angle in the triangle.

In this proof, angle C is the obtuse angle.

We drop a perpendicular from point B to intersect with the extension of side b at point D.

A picture of our triangle is shown below:

We are trying to prove that:

{{{c^2 = a^2 + b^2 - 2*a*b*cos(C)}}}

We start off with {{{c^2 = e^2 + h^2}}} from the right triangle ABD in the diagram. *****

***** means that this is the equation that will eventually become {{{c^2 = a^2 + b^2 - 2*a*b*sin(C)}}}.

This is from the well known Pythagorean Formula of {{{c^2 = a^2 + b^2}}} where:

Our c = Pythagoras' c
Our e = Pythagoras' a
Our h = Pythagoras' b

<img src = "http://theo.x10hosting.com/examples/LawOfCosines/Pythagorus.jpg" alt = "***** PICTURE NOT FOUND *****" />

From our diagram, we can see that:

{{{sin(BCD) = h/a}}}

From that, we get {{{h = a*sin(BCD)}}}

We can substitute in our equation of {{{c^2 = e^2 + h^2}}} to get:

{{{c^2 = e^2 + (a*sin(BCD))^2}}} *****

Since we know that the sine of the supplement of an angle is the same as the sine of the angle, we can substitute (C) for (BCD) to get:

{{{c^2 = e^2 + (a*sin(C))^2}}} *****

Since we know that {{{(a*sin(C))^2}}} is the same as {{{a^2*sin^2(C)}}}, we can substitute in our equation to get:

{{{c^2 = e^2 + a^2*sin^2(C)}}} *****

Since we can see from the diagram that {{{e = b+d}}}, we can substitute in our equation to get:

{{{c^2 = (b+d)^2 + a^2*sin^2(C)}}} *****

Since we know that {{{(b+d)^2 = b^2 + 2*b*d + d^2}}}, we can substitute in our equation to get:

{{{c^2 = b^2 + 2*b*d + d^2 + a^2*sin^2(C)}}} *****

Since we know that {{{cos(BCD) = d/a}}}, and from that we can derive {{{d = a*cos(BCD)}}}, we can substitute in our equation to get:

{{{c^2 = b^2 + 2*b*a*cos(BCD) + (a*cos(BCD))^2 + a^2*sin^2(C)}}} *****

Since we know that the cosine of the supplement of an angle equals minus the cosine of the angle, we can substitute -cos(C) for cos(BCD) to get:

{{{c^2 = b^2 + 2*b*a*(-cos(C)) + (a*(-cos(C)))^2 + a^2*sin^2(C)}}} *****

Since we know that {{{2*b*a*(-cos(C))}}} = {{{-2*a*b*cos(C))}}}, and since we know that {{{(a*(-cos(C)))^2}}} is the same as {{{a^2*cos^2(C)}}}, then we can substitute in our equation to get:

{{{c^2 = b^2 - 2*a*b*cos(C) + a^2*cos^2(C) + a^2*sin^2(C)}}} *****

Since we know that {{{a^2*cos^2(C) + a^2*sin^2(C)}}} = {{{a^2 * (cos^2(C) + sin^2(C))}}}, and since we know that cos2C + sin2C = 1, then we can substitute {{{a^2}}} for {{{a^2*cos^2(C) + a^2*sin^2(C)}}} in our equation to get:

{{{c^2 = b^2 - 2*a*b*cos(C) + a^2}}} *****

We can re-arrange the terms in our equation to get:

{{{c^2 = a^2 + b^2 - 2*a*b*cos(C)}}} *****

Since this is exactly the same as our original equation that we are trying to prove, then we are done.

This concludes the proof for Case 3.

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