Use one or more of the six sum and difference identities to solve...

Question

Use one or more of the six sum and difference identities to solve Exercises 13–54.In Exercises 13–24, find the exact value of each expression.sin 75°

Accepted Answer

Welcome back. I am so glad you're here. We are asked to determine the exact value of the following expression by using a sum or different identity. Our expression is the sign of 645 degrees. Our answer choices are answer choice. A negative quantity of square red of six plus squared of two divided by four answer choice B the quantity of the squared of six minus the square of two divided by four. Answer choice C the square of three divided by four and answer choice D, the negative quantity of square root of three plus square root of two divided by two. All right. So we've got the sign of 645 degrees. Well, that's more than a full rotation. That's more than 360 degrees. So we can find a co terminal angle with it, we can subtract 360 degrees. Oops that is there zero and then degrees. And when we subtract 360 degrees, we get a co terminal angle of 285 degrees. Now that one's not on our unit circle, which is why we want to use a sum or different identity to find out uh what the exact value would be. And we've got so many options. But if we can pull out a pretty easy reference angle from that, that can help. So we know that if we took the sign of pulling out, let's say 45 degrees from that, if we took 45 degrees, and then we could add 240 degrees that would bring us back up to our 285 degrees. And there we've got two angles that are both on the unit circle. What we can do is find the exact values of those two angles. And then we plug those in to, we can use a sum identity or a different identity and then we'll get the exact value. So what we'll do here is we'll use the sign of, we'll use our sum identity. We'll use the sign of A plus B which is equal to the sign of a multiplied by the cosine of B plus the cosine of a multiplied by the sign of B. Am I on the page? Just barely. Let's move that a little bit. All right. So there's the sum identity that we'll use. And so now let's figure out the exact values of the sign of 45 degrees and the sign of 240 degrees. I'll draw a rough sketch of a unit circle and we've got a horizontal X axis, a vertical Y axis. They come together the origin in the middle, overlaying that with a really rough unit circle, each quadrant has three hash marks on the unit circle. And so we know one of our angles is 45 degrees that is halfway between the positive parts of the X axis and the positive parts of the Y axis in the first quadrant. And so talking about the sign of 45 degrees, we can draw a rough sketch of a 45 to 45 90 triangle. So 45 degrees to the two angles that aren't our right angle at 90 degrees, the hypotenuse measures at the square root of two and the other two side lengths are both one. And so we're in the first quadrant with our 45 degrees sign is positive there. For the sign we are taking opposite, divided by the hypotenuse and the opposite here is one that's divided by the hypotenuse, which is the square of two. We don't want the radical in the denominator. So we'll rationalize it multiplying both the numerator and the denominator by the square root of two. And we'll get the square root of two divided by two. So the sign of 45 degrees, which is our angle. A here is the square root of two divided by two positive because that one is in the first quadrant. Now, let's find the cosine of 45 degrees. Well, here, the, the only difference is for the cosine of degrees, we'd be taking the adjacent divided by the hypotenuse, but the adjacent is also one hypotenuse is still the square of two. So this is also going to be the square root of two divided by two and still positive because we're still in the first quadrant. So that is also going to be the cosine of our angle A. So we can fill those in for the sign of 45 degrees plus 240 degrees. We know that the sign of angle A is the square root of two divided by two. Still need to multiply that by the cosine of angle B, we'll find that in a moment, adding that to the cosine of angle A will also be the square root of two divided by two. So now let's work on angle B. Angle B is our 240 degrees. And so we will be looking for 240 degrees on the unit circle and 240 degrees is in the third quadrant. And it's the hashmark that's closer to the negative part of the Y axis than the negative part of the X axis. And recalling from previous lessons, the reference angle for 240 degrees is 60 degrees. So we can draw a rough sketch of a 30 60 90 triangle. Mine conspicuously looks a lot like my 45 90 but I'll label it differently. So the smallest angle is 30 degrees. The right angle is 90 degrees and the one that's left is 60 degrees. The side lengths, if we're talking about 60 degrees, the opposite is going to be the square of three, the adjacent to 60 degrees is one and the hypotenuse is two. So for the sign of our reference angle, 60 degrees, again, we're taking the opposite divided by the hypotenuse and the opposite, as we said is the square of three, the hypotenuse is two, but we are in the third quadrant is sign positive or negative in the third quadrant and it's negative. So for the sign, we have got the sign of 240 degrees is going to be a negative square, it of three divided by two. And we can put that in that's being multiplied by our square root of two divided by two. We've got the negative square it of three divided by two. And now let's figure out our last one. So we want to find the cosine of 240 degrees. And again, our reference angle is 60. So we're finding the cosine of 60 degrees. Now we're taking the adjacent, divided by the hypo and the adjacent to 60 degrees is one and the hypotenuse is still two. We're still in quadrant three with our 240 degrees. So that is still negative. We've got negative one half is the cosine of degrees that is being multiplied by our square of two divided by two. OK. So we've got this all filled out for our sum identity. Now we just need to do the multiplication and the addition. So let's start off by multiplying the square it of two divided by two by negative one half. That's going to be a negative square it of two divided by four that's added to. Now, the second product we're taking the square root of two divided by two multiplied by the negative square root of three, divided by two. That's going to be a negative square rot of six divided by four. Now we have a common denominator. So we can combine our numerators, but they really can't be combined. They're not like terms, we can factor in negative out in front and then we can have negative quantity of square root of two plus the square root of six. And that's all divided by four. And that's our exact value because of the commutative property of addition. Our answer choice has in the quantity of in the numerator, the squared of six plus the squared of two instead of the square root of two plus the squared of six, but it's the same. So this matches with our answer choice. A well done. We'll catch you on the next one.

Use one or more of the six sum and difference identities to solve Exercises 13–54.In Exercises 13–24, find the exact value of each expression.sin 75°

Key Concepts

Sum and Difference Identities

Exact Values of Trigonometric Functions

Angle Decomposition

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