WEBVTT mathematics/math-analysis/selhorst-jones
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Hello, this is the trigonometry lectures for educator.com and today we're going to learn about probably the single most important identity in all trigonometry which is the Pythagorean identity.
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It says that sin²x + cos²x = 1.
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This is known as the Pythagorean identity.
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It takes its name from the Pythagorean theorem which you probably already heard of.
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The Pythagorean theorem says that if you have a right triangle, very important that one of the angles be a right angle, then the side lengths satisfy a² + b² = c².
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You probably heard that already.
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The new fact for trigonometry class is that sin²x + cos²x = 1.
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What we're going to learn is we work through the exercises for these lectures.
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Is if these are really two different sides of the same coin, you should think of this as being sort of facts that come out of each other.
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In fact, we're going to use each one of these facts to prove the other one.
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These are really equivalent to each other.
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Let's go ahead and start doing that.
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In our first example, we are going to start with the Pythagorean theorem, remember that's a² + b² = c².
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We're going to try to prove the Pythagorean identity sin²x + cos²x = 1.
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The way we'll do that is let x be an angle.
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Let's draw x on the unit circle.
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The reason I'm drawing it on the unit circle is because remember the definition of sine and cosine is the x and y coordinates of that angle.
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If we draw x on the unit circle, the hypotenuse has length 1 and the x-coordinate of that point, remember, is the cos(x), and the y-coordinate is the sin(x).
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Now, what we have here is a right triangle and we're allowed to use the Pythagorean theorem, we're given that and we're going to use that and try to prove the Pythagorean identity.
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The Pythagorean theorem says that in a right triangle, by the Pythagorean theorem...
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Let me draw my right triangle a little bigger, there's x, there's 1, this is cosx, this is sinx.
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By the Pythagorean theorem, one side squared, let me write that first of all as cosine x squared plus the other side squared is equal to 1².
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That's the length of the hypotenuse.
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If we just do a little semantic cleaning up here, 1², of course, is just 1, cosine x squared, the common notation for that is cos²x + sin²x = 1.
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We just derived an equation, and look this is the Pythagorean identity.
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What we've done is we started by assuming the Pythagorean theorem and then we used the Pythagorean theorem to derive the Pythagorean identity.
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Let's see an application of that in the next example.
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We're given that θ is an angle whose cosine is 0.47, and θ is in the fourth quadrant.
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We have to find sinθ.
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Let me draw θ, θ is somewhere down there in the fourth quadrant.
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I don't know exactly where it is but θ looks like that.
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Here is what I know, by the Pythagorean identity, sin²θ + cos²θ = 1.
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I'm going to fill in the one that I know, cosθ, cosθ is 0.47.
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This is 0.47² = 1 + sin²θ.
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Now, 0.47, that's not something I can easily find the square of, so I'll do that on my calculator.
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0.47² = 0.2209, so that's +0.2209, sin²θ +0.2209 = 1, sin²θ = 1 - 0.2209, which is 0.7791.
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Sinθ, if we take the square root of both sides, sinθ is equal to plus or minus the square root of 0.7791, which is approximately equal to 0.8827.
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Now, it's plus or minus because I know that sine squared is this positive number, but I don't know whether this sine is a positive or negative.
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We're given more information in the problem, θ is in the fourth quadrant.
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Remember, sine is the y-coordinate, so the sine in the fourth quadrant is going to be negative because the y-coordinate is negative.
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Because θ is in quadrant 4, sinθ is going to be negative, so we take the negative value, sinθ is approximately equal to -0.8827.
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The whole key to doing this problem was to start with the Pythagorean identity sin²θ + cos²θ = 1.
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Once you're given sine or cosine, you could plug those in and figure out the other one except that you can't figure out whether they're positive or negative.
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Their identity doesn't tell you that so we had to get this little extra information about θ being in the fourth quadrant, that totals that the sinθ is negative and we were able to figure out that it was -0.8827.
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Let's try another example of that.
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We're going to verify a trigonometric identity.
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This is a very common problem in trigonometry classes as you'll be given some kind of identity involving the trigonometric functions and you have to verify it.
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For this one, what I want to do is start with the right hand side.
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I'm going to label this RHS.
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RHS stands for right-hand side.
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The right-hand side here is equal to sinθ/(1 - cosθ).
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Now, I'm going to do a little trick here which is very common when you have something plus something in the denominator, or something minus something in the denominator.
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The trick is to multiply the conjugate of that thing.
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Here I have 1 - cosθ in the denominator, I'm going to multiply by 1 + cosθ, and then, of course, I have to multiply the numerator by the same thing, 1 + cosθ).
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The reason you do that, this is really an algebraic trick so you probably have learned about this in the algebra lectures.
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The reason you do that is you want to take advantage of this formula, (a + b) × (a - b².
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That's often the way of simplifying things using that algebraic formula.
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What we get here in the numerator is (sinθ) × (1 + cosθ), in the denominator, using this (a² + b²) formula, we get (1 - cos²θ).
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Now, let's remember the Pythagorean identity.
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Pythagorean identity says sin²θ + cos²θ = 1.
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That means 1 - cos²θ = sin²θ.
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We can substitute that in into our work here, sinθ×(1 + cosθ).
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The denominator, by the Pythagorean identity, turns into sin²θ.
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We get some cancellation going on, the sine in the numerator cancels with one of the sines in the denominator leaving us just with (1 + cosθ)/sinθ in the denominator.
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That's the same as the left-hand side that we started with.
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We started with the right-hand side and we're able to work it all the way down and end up with the left-hand side verifying the trigonometric identity.
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There were sort of two key steps there.
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One was in looking at the denominator and recognizing that it was a good candidate to invoke this algebraic trick where you multiply by the conjugate.
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If you have (a + b), you multiply by (a - b).
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If you have (a - b), you multiply by (a + b).
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Either way, you get to invoke this identity.
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Here, we had (a - b), we multiplied by (a + b) and then we got to invoke the identity and get something nice on the bottom.
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The second trick there was to remember the Pythagorean identity and notice that (1 - cos²θ) converts into sin²θ.
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Once we did that, it was pretty to simplify it down to the left-hand side of the original identity.
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We'll try some more examples later.