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### Course: High school geometry (staging)>Unit 6

Lesson 2: Law of cosines

# Proof of the law of cosines

Sal gives a simple proof of the Law of cosines. Created by Sal Khan.

## Want to join the conversation?

• Could you explain this using actual values instead of just letters? Like say you were given two sides and an angle?
• Refer to "Trigonometry word problems (part 1)"
• How is the sine of theta squared + the cosine of theta squared equal to one?
• sin x = {opposite side / hypotenuse }
sin^2(x) = [ opposite side / hypotenuse ]^2

cos x = {adjacent side / hypotenuse }
cos^2(x) = [ adjacent side / hypotenuse ]^2

sin^2(x) + cos^2(x) = [ opposite side / hypotenuse ]^2 + [ adjacent side / hypotenuse ]^2
= [ (opposite side)^2 + (adjacent side)^2 ] /( hypotenuse )^2
= (hypotenuse)^2 / (hypotenuse)^2
= 1
• Does anyone have a good link for practice problems for the law of cosines (preferably with the answers) ?
• You can make up your own problems! Cut a triangle out of a piece of paper, measure two sides and an angle, and see how closely you can predict the third side, or measure the three sides and see how closely you can predict the measure of any of the angles.
• how do you know when to use the:
the sine rule
cosine rule
tangent rule.
• Sine rule: When you have all the angles and a side, to calculate the other sides. (If you use it the other way, you will find two possible values for the angles, as sin( 80º ) = sin( 100º ), for example.)
Cosine rule: When you have the three sides and want to calculate an angle, or when you have two sides and an angle, and want to find the third side.
Tangent rule: When you want to be fancy.
• Why is (at ) it that (c-bcosQ)^2 = c^2 - 2cbcosQ + b^2cos^2Q ? I understand that he multiplied it out but I don't see how he got to that. Can someone please write out the logic?
• ( c - bCosQ ) ^ 2
= (c - bCosQ) x (c - bCosQ)
= (c x c) - (c x bCosQ) - (bCosQ x c) + (bCosQ x bCosQ)
= c^2 - cbCosQ - cbCosQ + b^2Cos^2Q
= c^2 - 2cbCosQ + b^2Cos^2Q
• how do you solve for side "a" if the only measurements you have are angle theta, the angle between sides "b" and "a" (the one from which you dropped the altitude) and the length of side "c"?
• how do you figure out which sides are a,b, and c?
• To solve for a, just use your known sides (doesn't matter which) for b and c, and the angle between them to be theta. Then just plug-and-chug the formula to find your unknown. If you are given three sides and need to solve for the angle, make sure the side you plug in for a is opposite the angle you want to find.
Note that some forms of the law solve for c using angle gamma or some other arrangement. For whatever form you are using, just dump the known sides on one side and solve for the unknown on the other, or put the side opposite the unknown angle by itself and dump the other sides into the rest of the function.
• Could this be applied along with the law of sines to make a law of tangents?
• How do you prove sin^2θ + cos^2θ = 1 ?
• There are many ways to do this. Here is a link that shows several of those proofs:
https://proofwiki.org/wiki/Sum_of_Squares_of_Sine_and_Cosine
I personally prefer the exponential formulation proof because it doesn't invoke any trig identities other than knowing what sine and cosine are in terms of e.
• Not a question - but a comment (smile). The "age" comments were cute -- I did take trig 5 years ago - just prior to retiring from the Army and beginning teaching of math. Now at 58 (a long cry after 40) I'm still doing math, and having to relearn Trig. So - there are some of us after age 40 still doing trig - smile
• Good for you - I am the same age, but Math stuck better in my head, so after a long stint working at the patent office, I went back to teaching high school students math, but most if what I have done so far I did not loose, but whenever I have to help a student with Calculus, they have to help remind me what to do, or I have to do the same refreshing of the mind.