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.

( 5𝝅 𝝅 )
tan ( ------ ﹣ ------)
( 3 4 )

Accepted Answer

Welcome back. I am so glad you're here. We're asked to determine the exact value of the following expression by using the difference identity. Our expression is the tangent of 11 pi divided by six minus three pi divided by four. Our answer choices are answer choice A, the quantity of two minus square at three divided by three answer choice B, the squarer of three divided by three answer choice C, the quantity of one minus square at two divided by three and answer choice D two minus square at three. All right. So what is the difference identity? We recall from previous lessons that that is when we are taking the tangent of two angles that are being subtracted. That's the tangent of, we'll use angle A minus angle B which is equal to in the numerator, the tangent of A minus the tangent of B all divided by, in the denominator one plus the tangent of A multiplied by the tangent of B. And so what we can do here first identify our angle A and angle B. So angle A is going to be the first one here in parentheses. So A equals pi divided by six and angle B, that's going to be the second one in parentheses that is three pi divided by four. What we want to do is identify the exact values of the tangents of both of those angles. So the tangent of 11 pi divided by six and the tangent of three pi divided by four. And then we want to put those exact values in to our difference identity equation then simplify. So let's start off with the exact value for the tangent of pi divided by six. We recall from previous lessons, the unit circle, we have a vertical y axis, a horizontal X axis. They come together at the origin in the middle. We overlay this with the really rough sketch of a unit circle. 11 pi divided by six is in the fourth quadrant that is closer to the positive part of the X axis, then the negative part of the y axis. And so we're talking about a quadrant four angle. Let's figure out what that reference angle is. The reference angle. We would take two pi minus our quadrant four angle 11 pi divided by six, which would give us pi divided by six as our reference angle. And we recall from previous lessons and the tables in the book that the tangent of our reference angle pi divided by six is the square of three divided by three. Now, the only difference is we're talking about a quadrant four angle with 11 pi divided by six is tangent positive or negative and quadrant four, it is negative. So for the tangent of 11 pi divided by six, that's negative square at three divided by three. So we found that exact value. Now let's do three pi the tangent of three pi divided by four, all right, three pi divided by four is in quadrant two. That's exactly halfway between the positive part of the y axis and the negative part of the X axis. So for that quadrant two angle, our reference angle finding that for quadrant one is going to be pi minus three pi divided by four which will give us pi divided by four. So for the tangent of pi divided by four, we recall from previous lessons in the tables in the book that that is going to be a positive one. But the tangent of three pi divided by four is in the second quadrant is tangent positive or negative. In the quadrant two, it is negative. So the tangent of three pi divided by four's exact value is negative one. Now we can plug in negative square at three divided by three. And for the tangent of angle A and we can plug in negative one for the tangent of angle B. So let's do that. So in the numerator tangent of angle A is negative square at three divided by three and minus the tangent of angle B which is negative one and that's all divided by in the denominator one plus tangent of angle A is negative square at three divided by three multiplied by tangent of angle B which is negative one. Now, this is just a lot of simplifying and it's actually a lot of steps to simplify this particular one. So a few things we can look at first in the numerator, we are subtracting a negative with our minus negative one. Well, that's just going to simplify to a plus one. And we want to add a one but the negative square at three divided by three has a denominator of three. So we'll change that one to three thirds. So this will simplify to be in the numerator negative square at three plus three all divided by three. Now I'm going to put a division symbol and we'll put the denominator next to it instead of below it with fractions upon fractions. So let's simplify the denominator. We've got another negative multiplied by a negative. So those are going to be positives. And then again, we have a one being added to something with the denominator of three. So we'll multiply that one by three thirds. So we have three thirds plus, well, one multiplied by the square root of three, divided by three is the square root of three divided by three. So three thirds plus square at three, divided by three is going to be in the new writer three plus square at three, all divided by three. Now we know that when we are dividing two fractions, we can multiply by the second fractions reciprocal. So let's do that. We'll have negative square at three plus three divided by three, multiplied by three, divided by three plus square at three. Now, these threes that are hanging out by themselves as factors are going to cancel out. Those are ones. So this simplifies to B negative square at three plus three divided by three plus square at three. Now, we need to rationalize this denominator. We do not want the square out of three in our denominator. So we're going to multiply it by its complex conjugate, which would be the A term which is three minus the B term square at three. And what we do to the denominator we do to the numerator as well multiplying both by three minus square at three. So in the numerator, we want to foil that. So taking negative square at three, multiplied by three is going to give us three or excuse me, negative three square at three. And then in our outsides negative square at three multiplied by negative square at three, that one's going to be a plus three for our insides three multiplied by three. And that's going to be plus nine. And then for our last three multiplied by negative square at three is going to be minus three square at three. So there's our new writer not simplified yet. And for our denominator, we recall from previous lessons that when we multiply a complex number by its complex conjugate, we are going to get A squared minus B squared. Our A, as we said is three, so three squared, that's nine and our B as we said is square at three square at three squared is just three. So we'll have nine minus three. Now, let's combine like terms in the numerator three plus nine is 12. a negative square at three or negative three square at three minus three square at three is negative six square at three. That's all divided by nine minus three, which is six. In the numerator, we can factor out a six. So we'll have six multiplied by the quantity of two minus square at three, all divided by six. Those six is canceled out. And so we have our final answer of two minus square at three and we look at our answer choices and that matches with answer choice. D well done. We'll catch you on the next one.

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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. ( 5𝝅 𝝅 ) tan ( ------ ﹣ ------) ( 3 4 )