So it's past time for another math post. Don't worry, my studies are continuing, but I've discovered that distilling all the great mathy things I'm learning into a coherent blog post is actually more difficult than studying the math! Still, that's the reason for keeping the blog - if I can't explain what I'm learning in simple words, I don't truly understand it.
So - it's taken me a little while to prepare a post on the Law of Cosines. Mostly because it took me a little while to wrap my brain around it - but a little bit of lightning struck my brain and I suddenly got it. Here's the gist:
So - it's taken me a little while to prepare a post on the Law of Cosines. Mostly because it took me a little while to wrap my brain around it - but a little bit of lightning struck my brain and I suddenly got it. Here's the gist:
Here's a triangle. It has sides of length a, b, and c, and angles which measure A, B, and C.
Now - note that there are no right angles here, which makes this triangle different from any of the other ones we've used so far. The law of cosines and the law of sines are used to determine the relationships between the angles and sides of these triangles, and frankly I think the law of cosines is the more useful of the two, but I'll show you both.
Now, for a right triangle we can calculate the relationship between its sides pretty simply, using the Pythagorean theorem. The Law of Cosines, in fact, is just an extension of that which will work for any triangle. Here it is:
Now - note that there are no right angles here, which makes this triangle different from any of the other ones we've used so far. The law of cosines and the law of sines are used to determine the relationships between the angles and sides of these triangles, and frankly I think the law of cosines is the more useful of the two, but I'll show you both.
Now, for a right triangle we can calculate the relationship between its sides pretty simply, using the Pythagorean theorem. The Law of Cosines, in fact, is just an extension of that which will work for any triangle. Here it is:
Now. I have been through the proof for this equation a couple of times, and while I appreciate it, I sort of resolved just to smile and know that this is true. The lightning-bolt moment came when I tried to apply the law of cosines to a right triangle, where C=90 degrees. The cosine of any 90 degree angle is the same as the x coordinate of that radius' endpoint on the unit circle, remember - so, it's zero. Plug that into the equation above and it becomes, quite beautifully, the Pythagorean theorem.
It also works for finding the length of any of the three sides, not just c, as long as you replace the angle after "cos" with the angle opposite the side you're trying to find.
In Practice
So let's give this a shot. Back to our original triangle, let's assign some values.
It also works for finding the length of any of the three sides, not just c, as long as you replace the angle after "cos" with the angle opposite the side you're trying to find.
In Practice
So let's give this a shot. Back to our original triangle, let's assign some values.
a = 8
c = 11
B = 37 degrees.
c = 11
B = 37 degrees.
So we know the lengths of two sides, and the measure of one angle. Let's see if we can use that to find out the length of side B. We'll plug the values we know into the Law of Cosines (with apologies for the lack of superscript, I just can't be bothered to create images or html for all this crap. :P
b^2 = a^2 + c^2 - 2ab cos(B)
b^2 = 8^2 + 11^2 - 2 8 11 cos(37 deg)
b^2 = 64 + 121 - 176 cos(37deg)
b^2 = 185 - 176 cos(37deg)
We'll have to use a calculator to find the cosine of 37 degrees, or look it up. Here it is.
cos (37 degrees) = 0.798...
so...
b^2 = 185 - 176 (0.798...)
b^2 = 185 - 140.559...
b^2 = 44.440...
b = sqrt(44.440...)
b = 6.666...
Wow. I promise I didn't plan that.
I did use a problem that I had the answer to, just to verify that I did it correctly but I didn't look at the solution until I was done. Success! Now we can find the other two angles, if we want. If you're looking for the angle, you just have to reconfigure the equation solving for cos(x).
Thus:
c^2 = a^2 + b^2 - 2ab cos(C)
c^2 - a^2 - b^2 = -2ab cos(C)
-c^2 + a^2 + b^2 = 2ab cos(C)
(-c^2 + a^2 + b^2) / 2ab = cos (C)
There you have it. Yay algebra! Now, here's the less useful Law of Sines.
The Law of Sines
The Law of SInes is simply a statement of the ratios between the angles and their opposite sides. So, in our same triangle from before:
b^2 = a^2 + c^2 - 2ab cos(B)
b^2 = 8^2 + 11^2 - 2 8 11 cos(37 deg)
b^2 = 64 + 121 - 176 cos(37deg)
b^2 = 185 - 176 cos(37deg)
We'll have to use a calculator to find the cosine of 37 degrees, or look it up. Here it is.
cos (37 degrees) = 0.798...
so...
b^2 = 185 - 176 (0.798...)
b^2 = 185 - 140.559...
b^2 = 44.440...
b = sqrt(44.440...)
b = 6.666...
Wow. I promise I didn't plan that.
I did use a problem that I had the answer to, just to verify that I did it correctly but I didn't look at the solution until I was done. Success! Now we can find the other two angles, if we want. If you're looking for the angle, you just have to reconfigure the equation solving for cos(x).
Thus:
c^2 = a^2 + b^2 - 2ab cos(C)
c^2 - a^2 - b^2 = -2ab cos(C)
-c^2 + a^2 + b^2 = 2ab cos(C)
(-c^2 + a^2 + b^2) / 2ab = cos (C)
There you have it. Yay algebra! Now, here's the less useful Law of Sines.
The Law of Sines
The Law of SInes is simply a statement of the ratios between the angles and their opposite sides. So, in our same triangle from before:
Simply put, the ratio between the length of side a and the sine of A is always the same as the ratio of the length of side b and the sine of B, and so on. Useful to know, but I gather the Law of Cosines is used more often. I've been through the proof of this rule as well - I think it's useful to understand why these things are true and not just accept that they are.
The Quadratic Surprise
I had forgotten completely about the quadratic formula. Imagine my surprise when I stumbled upon a trig problem that seemed just unsolveable at first because I couldn't isolate the variable.... when I looked up the solution, I saw that you had to rephrase the equation as a quadratic first. I don't remember the details of the problem but a quadratic equation, basically, looks like this:
ax^2 + bx + c = 0
Where a, b, and c are given, and are known as the quadratic coefficients. Then you use the quadratic formula to solve for x:
The Quadratic Surprise
I had forgotten completely about the quadratic formula. Imagine my surprise when I stumbled upon a trig problem that seemed just unsolveable at first because I couldn't isolate the variable.... when I looked up the solution, I saw that you had to rephrase the equation as a quadratic first. I don't remember the details of the problem but a quadratic equation, basically, looks like this:
ax^2 + bx + c = 0
Where a, b, and c are given, and are known as the quadratic coefficients. Then you use the quadratic formula to solve for x:
From there, just plug in and solve. I would never have gotten here by myself, though, because I didn't immediately recognize the problem as quadratic. When I first started this endeavor, someone told me that when they get to calculus, most students' major handicap isn't geometry or trigonometry, it's algebra. I can see here why. The Quadratic Formula was something we talked about in Algebra 2 (I think?) that I hadn't even though about since. I'm pretty sure I'll have to go back and review algebraic ideas like this from time to time. Is it strange that, rather than getting frustrated, I regarded it as a nice surprise? Probably is... but I am having fun, and the major upshot of teaching myself this stuff is that I have time go back and review older ideas like this, because I don't have to stick to a schedule, and I'm not depending upon a grade. I'm very thankful for that!
Until next time...
Until next time...
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