Today, I am just going to focus on the Law of Cosines, and save the Law of Sines for a future post. This formula is one where you probably wondered why it worked. You might be asked to prove something using this formula, but how can you comfortably do that without being sure of the formula in the first place?

First, I need to lay a little foundation. Take a right triangle:

First off, we should know the Pythagorean Theorem. Click here for an explanation and proof of it.

Let's say we are dealing with the bottom left angle. The side opposite to this angle has a measure of 4, and will be referred to as the "opposite" side. The longest side has a measure of 5, and will be referred to as the "hypotenuse." The remaining side has a measure of 3, and will be referred to as the "adjacent" side, since it is adjacent to the angle.

Let's say we are dealing with the bottom left angle. The side opposite to this angle has a measure of 4, and will be referred to as the "opposite" side. The longest side has a measure of 5, and will be referred to as the "hypotenuse." The remaining side has a measure of 3, and will be referred to as the "adjacent" side, since it is adjacent to the angle.

Now, let's make sure we are on the same page with terminology. The sine of this angle is the opposite side over the hypotenuse. 4/5 is 0.8, so the sine of that angle is 0.8. The cosine of this angle is the adjacent side over the hypotenuse. 3/5 is 0.6, so the cosine of that angle is 0.6. Tangents will not be needed in this post, but it is the opposite over the adjacent. These ratios are commonly remembered by the SOH CAH TOA acronym.

There is a pretty cool identity found in the sine and cosine ratios. Take the two ratios that we just found for the 3-4-5 triangle:

sin(

*x*) = 0.8
cos(

*x*) = 0.6
Now, plug those into the following expression:

sin

0.8

0.64 + 0.36

1

Interestingly, this sum always turns out to be one. By using those ratios and the Pythagorean Theorem, you can prove that for all angles, the square of the sine plus the square of the cosine is one.

With this information, we can prove the Law of Cosines. The formula slightly resembles the Pythagorean Theorem, so we will try to keep that in mind when proving it.

^{2}(*x*) + cos^{2}(*x*)0.8

^{2}+ 0.6^{2}0.64 + 0.36

1

Interestingly, this sum always turns out to be one. By using those ratios and the Pythagorean Theorem, you can prove that for all angles, the square of the sine plus the square of the cosine is one.

With this information, we can prove the Law of Cosines. The formula slightly resembles the Pythagorean Theorem, so we will try to keep that in mind when proving it.

Each of the derived measurements are just rewritten forms of the sine and cosine ratios. For example, the cosine of C is that adjacent side over b, so multiplying both sides by b yields the measure of the adjacent side: b • cosC.

Let's look at the right triangle with hypotenuse c, and solve the Pythagorean Theorem. Since the Law of Cosines is similar to the Pythagorean Theorem, this might give us a start.

[a - (b • cosC)]

^{2}+ (b • sinC)

^{2}= c

^{2}

a

^{2}- 2abcosC + b

^{2}cos

^{2}C + b

^{2}sin

^{2}C = c

^{2}

a

^{2}+ b

^{2}cos

^{2}C + b

^{2}sin

^{2}C - 2abcosC = c

^{2}

a

^{2}+ b

^{2}(cos

^{2}C + sin

^{2}C) - 2abcosC = c

^{2}

^{}Wait a minute - what is inside the parentheses? We have cos

^{2}C + sin

^{2}C. But we proved earlier that that is always equal to one. So, we can substitute one in for that sum, and see where that takes us.

a

^{2}+ b

^{2}(cos

^{2}C + sin

^{2}C) - 2abcosC = c

^{2}

a

^{2}+ b

^{2}(1) - 2abcosC = c

^{2}

a

^{2}+ b

^{2}- 2abcosC = c

^{2}

^{}And we end up with the Law of Cosines: a

^{2}+ b

^{2}- 2abcosC = c

^{2}. I think that this proof is a pretty cool one, considering that it brings so many other neat identities into play.

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