Hey everybody, this is lesson 2 in unit 5. We're going to look at proving trig identities. Before we get started, here are a few guidelines to verifying trig identities that'll be helpful. We always want to pick one side of the equation to work with, and usually it's easier to start with the more complicated side and change that side to look identical to the other side. You can change every expression to be in terms of either sine or cosine. And then a third thing you can do is to look for algebraic operations like adding fractions, factoring, distributing, squaring terms, multiplying by conjugates, multiplying terms together, or use some kind of clever form of a 1. So those are just some tips and tricks we can use.
You can also look for trig things to do, like if there's squares of functions, you could apply the Pythagorean identities. And remember things like cosine and secant is 1, or sine times cosecant is 1, and tangent times cotangent is 1. This is always true. You keep looking at your answer, that's your goal, to make sure you're on the right path.
And remember, there will often be more than one way that you can verify an identity, so don't quit. If one idea doesn't work, then go back and try something else. And when you're finished, if you like, you can put QED at the end of the proof that stands for quo erat demonstratum, which was to be demonstrated.
So it's just a mathy way of saying you finished. So first we're going to start off with proving an algebraic identity. And you notice that they have different denominators. So to combine like terms, you would want to get both sides having a common denominator.
I'm also using the idea that I'm factoring the numerator. And by factoring the numerator on both sides, Oop, messed that up. You can see we can eliminate the denominators in both cases.
So you have x plus 1 and x minus 1 since these are the difference of two squares, and we want to know if it's equal to 2. So in the first fraction, you see we have the common x minus 1. And over here we have x plus 1, so simplifying that we have x plus 1 minus x minus 1, and I still want to know is that equal to 2. So dropping the parentheses and distributing the subtraction, you can see that, um, This x and this x go away, and 1 plus 1 is 2. And yes, 2 equals 2. So we have demonstrated that which was to be demonstrated, and we've proven that identity. So that was an algebraic approach. In the second example, we want to work with the left side, unless the right side is obviously more complicated.
And here you can see by looking at it that the left side is the more complicated. The quotient reciprocal identity here for cosine squared on the bottom could be rewritten as secant squared on the top. And secant squared minus 1 is one of our Pythagorean identities. So secant squared minus 1 is another name for tangent squared. And tangent squared is equal to the right hand side.
And we have proven or demonstrated that which was asked to be demonstrated. In this case, I want to turn tangent in terms of sine and cosine by using the quotient property. And doing the same thing with cotangent, I could say that's cosine.
divided by sine, and we don't want to mess with the right hand side, so I want to prove that the left hand side, which is the sum of two terms, can be turned into the product of two terms. So in this case, I want to get a common denominator. The common denominator would be sine times cosine, and in order to do that, I would multiply this fraction by sine over sine, so sine squared on the top.
And this fraction, I would multiply this by cosine over cosine so that it could have this denominator. And here, In the numerator, you see we have a 1, so sine, cosine on the bottom, and I'm trying to keep this side the same. And I can see if I rewrite that as 1 over sine times 1 over cosine and rewrite 1 over sine as cosecant.
And 1 over cosine as secant, then I do have secant times cosecant just by moving the two terms. Since it's a multiplication problem, they can commute and we are definitely finished. So we've proven that the left side is definitely equal to the right side. Example 4, in this one we have an addition problem and we want to match it to one of these two choices. So again, I need to have a common denominator.
And my well-chosen one would be a secant x plus 1 on the left. That would be a form of 1. And then the same on here, secant x minus 1 on the right would give us that. common denominator of secant x plus 1 times secant x minus 1. So using distributive property here and using distributive property here, we're going to have secant x plus 1 plus secant x minus 1. Combining like terms, we're going to get 2 secant x in the numerator and then secant squared x minus 1 in the denominator. But my denominator here, this is a Pythagorean identity, so using that truth, we just had that before, secant squared minus 1 is equivalent to tangent squared x.
Tangent squared x, and I'm looking back at my answer, so I believe it's going to turn out to be this one, because I can see in my work, if we separate secant and put 2 times secant as 2 divided by cosine, and then tangent squared on the bottom would be a cotangent. So that could be cosine squared in the top and sine squared in the bottom. So this cosine would eliminate one of the powers, and that's going to give us 2 and cosine divided by sine, and then use the other sine as 1 divided by.
Sine, this becomes cotangent, so 2 cotangent x, and this is cosecant x, so 2 cotangent x, cosecant x is choice A. Okay, so let me go back just, excuse me, just in case I confused you here. What I was thinking about is thinking this, that 2. Times secant and cotangent is how I ended up moving this to the top and secant moved to cosine in the bottom.
So we put secant on the bottom as cosine and cotangent on the top for the tangent. And that's how I got to this step. Okay? So that's example four.
And here's one in example five. We can maybe rewrite or decompose all of these. So we got sine squared.
No, we'll just change this. Here's that fact. Cosine times secant is a one, right? So we got sine squared here and cosine here and secant is one over cosine.
So this. Can remove that. We have sine squared, and I'm wondering is that 1 minus cosine squared? And we know by the Pythagorean identities, sine squared plus cosine squared is 1. So 1 minus cosine squared is also sine squared, and that one was pretty easy.
So we displayed and distributed, demonstrated what we needed. Okay, let's see. We've got, how can we prove that cosine over 1 minus sine is equal to 1 plus sine over cosine? So here I think I want to use a well-chosen one and create a Pythagorean identity on the bottom.
Okay, so if I use one plus sign on the left and don't touch the right, so we're trying to prove is the left equivalent to this expression on the right. So that's our question. So in the top I have, leaving that without distributing, I have one plus sign times cosine.
And then in the bottom, we've made an identity, 1 minus sine squared. And I'm going to just follow along here. And 1 minus sine squared on the bottom is cosine squared on the bottom.
But we have cosine in the numerator times 1 plus sine. So this power of 2 would eliminate that power in the numerator. And that's the answer we were looking for. 1 plus sine of t divided by cosine of t is the equivalent statement on the right.
And we have proven that which was to be demonstrated. So that's our last step there. Example 7 is a little tougher. So again, the left side is the more complicated side.
And I'm going to rewrite. I've got a 1 here, so I could rewrite that 1 as 1 plus cosine over 1 plus cosine. Because any number divided by itself is a 1. And that way, we have cleverly gotten A common denominator.
So we have 1 plus cosine is our common denominator. Is that messy looking fraction equivalent to cosine? Well, we have 1 plus cosine in the bottom, and 1 plus cosine minus sine squared, but sine squared is 1 minus cosine squared.
from our Pythagorean identity. And distributing the subtraction, we're adding a negative 1 and we're adding a cosine. So in this case, we have 1 minus 1 is eliminated, and we have a cosine plus a cosine squared.
and we're dividing by 1 plus cosine. Is all of that equivalent to cosine? So factoring out a cosine, we have a 1 plus cosine factor in the numerator, and 1 plus cosine factor in the denominator, and this is another form of 1, so any number divided by itself. can be removed and we have cosine Equal to cosine and we proved what needed to be proven.
I have two more examples. So cosecant minus sine, can we turn this subtraction problem into a product and show that it's equivalent to cotangent times cosine? So we'll do that by changing everything to sine and cosine.
Cosecant is 1 divided by sine, and we're subtracting sine. So I need a common denominator. I can give this sine over sine. For this fraction, and that would give us 1 minus sine, 1 over sine on the first fraction, and sine squared in the numerator over sine. Now we have two fractions with a common denominator.
But 1 minus sine squared is cosine squared over that common denominator. And I'm checking back at my answer up here because I want that to become a cotangent times a cosine. And I've got two cosines in the top.
So I can separate that, tear that fraction apart. Cosine over sine times another cosine. This is my cotangent times cosine, and that was where we were going. Cotangent times cosine.
So these are like little puzzles. They're fun and frustrating at the same time. Then our last one is a little more complicated. So I know that I've got a bunch of secants and cosecants.
Whoops, that's not right. Secant, cosecant, secant. Yeah, something's up with this.
That looks like a cut and paste error. Oh, now I see. The left side is multiplication, but the right side is addition. Okay, so to get out of that, let's turn secant squared into a Pythagorean replacement.
So 1 plus tangent squared is secant squared, and we're multiplying times cosecant squared. And we can turn a multiplication problem into an addition problem by using that Pythagorean identity. So cosecant times one is cosecant squared, and tangent squared times cosecant squared. We can rewrite the tangent squared as sine over cosine.
So this is going to be tangent squared is sine squared over cosine squared and the cosecant squared. can become sine squared in the denominator. So we took the tangent times the cosecant and rewrote it to look like this. So we can remove the sine squares.
and end up with cosecant squared plus one over cosine squared, but that's what we wanted up here, right? And cosine squared on the bottom. can be used as the reciprocal identity secant squared, and that's what we were looking for.
We were looking for secant squared plus cosecant squared, and it's an addition problem, so I addition their commutative. And secant squared plus cosecant squared is equal to the right-hand side. And that's the end of lesson two. See you next time.