WEBVTT mathematics/pre-calculus/selhorst-jones
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OK, welcome back to the trigonometry lectures on educator.com, and today we're talking about the identity, tan²(x)+1=sec²(x).
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You really want to think about this as a kind of companion identity to the main Pythagorean identity, which is that sin²(x)+cos²(x)=1.
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Hopefully, you've really memorized this identity, sin²(x)+cos²(x)=1, that's one that comes up everywhere in trigonometry.
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Then sort of the companion Pythagorean identity to that for secants and tangents is tan²(x)+1=sec²(x).
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There's another related identity, essentially just the same for cotangents and cosecants, is that cot²(x)+1=csc²(x).
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You may wonder how you'll remember this identities.
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For example, how do you remember whether it's tan²+1=sec², or maybe it's the other way around, sec²+1=tan².
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Well, really the answer to that lies in the exercises that we're about to do.
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We'll show you how to derive those identities from the original one sin²+cos²=1.
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As long as you can remember sin²+cos²=1, you'll learn how you can use that to figure out the others and you really won't have to work hard to remember this new Pythagorean identities.
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Let's jump right into that.
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The first example here is to prove the identity tan²(x)+1=sec²(x).
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The trick there is to remember the original Pythagorean identity, which is cos²(x)+sin²(x)=1, that's the original Pythagorean identity.
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That's one that you really should memorize and remember throughout all your work with trigonometry.
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What you do to manipulate this into the new identity, is you just divide both sides by cos²(x).
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On the left, we get cos²(x)/cos²(x) plus, let me write it as (sin(x)/cos(x)²=1/cos²(x).
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Then of course, cos²(x)/cos²(x) is just 1, sin/cos, remember that's the definition of tangent, so this is tan²(x), 1/cos, that's the definition of sec(x), so sec²(x).
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You can just rearrange this into tan²(x)+1=sec²(x).
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That shows that this new identity is just a straight consequence of the original Pythagorean identity, so really if you can remember that Pythagorean identity, you can pretty quickly work out this new Pythagorean identity for tangents and cotangents.
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Let's try using the Pythagorean identity for tangents and cotangents.
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In this problem, we're given the tan(θ)=-4.21, and θ is between π/2 and π, we want to find secθ.
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Remembering the Pythagorean identity, tan²(θ)+1=sec²(θ).
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If we plug in what we're given here, tan²(θ), that's (-4.21²)+1=sec²(θ).
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Well, 4.21², that's something I'm going to workout on my calculator, is 17.7241.
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So, 17.7241+1=sec²(θ), that's 18.7241=sec²(θ).
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If I take the square root of both sides, I get sec(θ) is equal to plus or minus the square root of 18.7241.
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Again, I'm going to do that square root on the calculator, and I get approximately 4.33.
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The question here is whether we want the positive or negative square root, and that's where we use the other piece of information given in the problem.
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θ is in the second quadrant here, θ is between π/2 and π, so θ is somewhere over here.
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Now, sec(θ) remember, is 1/cos(θ).
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In the second quadrant, if you remember All Students Take Calculus.
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In the second quadrant, sine is positive, but cosine is not, cosine is negative.
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That means, sec(θ) is negative.
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θ is in quadrant 2, cos(θ) is negative, sec(θ), because that's 1/(θ), is negative.
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So, sec(θ), we want to take the negative square root there, and we get an approximate value of -4.33.
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That negative is very important.
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The key to that problem is first of all invoking this Pythagorean identity, tan²+1=sec².
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That's very important to remember.
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We plug in the value that we're given and then we work it through and we'll try to solve for sec(θ).
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We get this plus or minus at the end because we don't know if we want the positive square root or the negative square root.
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That's where we use the fact that θ is in the second quadrant.
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Since θ is in the second quadrant, cos(θ) must be negative, sec(θ), remember, is just 1/cos(θ), must also be negative.
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That's how we know we want the negative square root there, so we get -4.33.
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In our next example, we're given trigonometric identity and we're asked to prove it, (csc(θ)-cot(θ))/(sec(θ)-1) = cot(θ).
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I'm going to start with the left-hand side of this trigonometric identity and I'm going to manipulate it.
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I'm going to try and manipulate it into the right-hand side.
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The left-hand side which is (csc(θ)-cot(θ))/(sec(θ)-1).
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Now, I'm going to do something clever here and it's based on this old principle of whenever you have an (a-b) in the denominator, try multiplying by the conjugate, (a+b)/(a+b).
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If it's the other way around, if you have (a+b), you multiply by the conjugate (a-b)/(a-b).
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The reason you do that is that it makes the denominator (a²-b²).
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We're taking advantage of that old formula from algebra, the difference of squares formula, and a lot of times the (a²-b²) is something that it'll simplify into something nice.
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That's what's going to happen in here.
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We're going to multiply, since we have (sec(θ)-1 in the denominator, by sec(θ)+1.
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What that turns into in the denominator is this (a²-b²) form, so sec²(θ)-1.
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The numerator really doesn't have anything very good happening yet, (csc(θ)-cot(θ))×sec(θ)+1, nothing very good happening in the numerator yet.
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In the denominator, we're going to take advantage of the fact that tan²(θ)+1=sec²(θ), that's the Pythagorean identity that we're studying today.
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If you move that around, you get sec²(θ)-1=tan²(θ).
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That converts the denominator into tan²(θ), so that's very useful.
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In the numerator, we have csc(θ)-cot(θ) and sec(θ)+1.
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I think now I'm going to translate everything into sines and cosines because those will be easier to understand than cosecants and cotangents.
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So, cosecant, remember is 1/sin(θ), minus cotangent is cos(θ)/sin(θ), sec(θ)+1, well, sec(θ) is 1/cos(θ), and tan(θ), if we translate into sines over cosines, that's sin²(θ)/cos²(θ).
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Now, I've got fractions divided by fractions, I think the easiest thing to do here is to bring the denominator, flip it up and multiply it by the numerator.
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We get cos²(θ)/sin²(θ), that's flipping up the denominator, and then the old numerator is, if I combined the first two terms of the first one, it's (1-cos(θ))/sin(θ).
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In the second one, we have 1/cos(θ)+1.
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Now, I think what I'm going to do is I'm going to look at this cosine squared, write it as cos(θ)×cos(θ).
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Then pull one of those cosines over and multiply it by this term, and try to simplify things a little bit that way.
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That will leave me with just one cosine left, still have sin² in the denominator, (1-cos(θ)/sin(θ)) times, now, cos(θ)×(1/cos(θ) is 1, plus 1×cos(θ).
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Now, look, we have (1-cos(θ))×(1+cos(θ)).
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We're going to use this pattern again, the (a-b)×(a+b)=a²-b².
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If cos(θ)/sin²(θ), now multiply (1-cos)×(1+cos) gives us (1-cos²(θ)) and sin(θ) in the denominator.
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Now, we're going to use the other Pythagorean identity, the original one which are right over here, sin²(θ)+cos²(θ)=1, if you switched that around 1-cos²(θ)=sin²(θ).
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I'll plug that in, cos(θ)/sin(θ), 1-cos²(θ), now translates into sin²(θ) all divided by sin(θ).
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I forgot my squared there, that's from the line above.
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Now, we have some cancellations, sin²(θ) cancel, we get cos(θ)/sin(θ), but that's cot(θ).
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By definition of cotangent, cotangent is defined to be cos/sin.
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That's exactly equal to the right-hand side of the identity.
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When you're proving this trigonometric identities, you pick one side and you multiply it as much as you can.
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You try to manipulate it into the other side.
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It does takes some trial and error.
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I worked this problem out ahead of time, I tried a couple of different things and I finally found a way that gets us through it relatively quickly.
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It's not like you'll always know exactly which path to follow, it takes a little bit of experimentation.
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There are some patterns that you see over and over again, and the two patterns that we really invoke strongly in this one are these algebra pattern where (a-b)×(a+b)=a²-b².
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Essentially, whenever you see an (a-b) or an (a+b), think about multiplying it by the conjugate, and then taking advantage of that pattern.
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That's what really got us started on the first step here.
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We had a sec(θ)-1, and I thought, okay, let's multiply that by sec(θ)+1, that gives us sec²-1.
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The second pattern that we use there was to invoke these Pythagorean identities, tan²+1=sec², and sin²+cos²=1.
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We invoked that one here, converting sec²-1 into tan².
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Then later on, when he had another (a-b)×(a+b), it converted into 1-cos² and that in turn, by the Pythagorean identity converted into sin².
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It's just a lot of manipulation but you can let your manipulation be kind of guided by these common algebraic identities and these common Pythagorean identities.
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Then you just try to keep manipulating until you get to the other side of the equation.
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We'll try some more examples later on.