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.

( 𝝅 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 a sum identity. Our expression is the tangent of seven pi divided by six plus five pi divided by three. Our answer choices are answer choice A, the square it of two divided by two answer trace B the negative square of three divided by two answer choice C one and answer choice D the negative squarer of three divided by three. All right. So we recall from previous lessons that the sum identity for a tangent of two angles added together is the tangent of A plus B equals in the numerator. The tangent of A plus the tangent of B all divided by, in the denominator one minus the tangent of A multiplied by the tangent of B. And so because we are looking for exact values what we can do first is find the exact value of the tangent of seven pi divided by six. And then we will be plugging that in for the tangent of A. And then we want to find the exact value of the tangent of five pi divided by three and we'll plug that in for the tangent of B and then we'll simplify. So let's begin working on the tangent of seven pi divided by six. So we will recall from previous lessons, a really rough sketch of a unit circle. We've got a horizontal X axis, a vertical y axis. They come together at the origin in the middle. We'll overlay this with a rough unit circle. Each quadrant has three hash marks. And we recall from previous lessons that seven pi divided by six is in the third quadrant and it is the hash mark that is closer to the negative part of the X axis than the negative part of the Y axis. And recalling from previous lessons, the reference angle for seven pi divided by six is going to be 30 degrees. So talking about the reference or excuse me, talking about the tangent of 30 degrees, we can draw a rough sketch of a 30 60 90 triangle, 30 degrees being the short, the smaller angle, 90 degrees being the right angle and 60 degrees being the one that's left. Now labeling our sides opposite of 30 degrees. The length is one adjacent to 30 degrees is the square of three and the hypotenuse is two. So for the tangent of 30 degrees, we want opposite divided by adjacent, opposite as we said is one divided by the adjacent as the square of three. But we don't want a radical in the denominator. So we're going to rationalize this multiplying both the numerator and the denominator by the square of three. And so in the numerator, one multiplied by the square of three is the square it of three divided by, in the denominator, the squared of three multiplied by the squared of three is just three. But one last very important question where in the third quadrant is tangent positive or negative in the third quadrant and tangent is positive in the third quadrant. So the exact value of the tangent of seven pi divided by six is the square, it of three divided by three. Now let's work on finding the exact value of the tangent of five pi divided by three. So looking at our unit circle, five pi divided by three is in the fourth quadrant and it is the hashmark that's closer to the negative part of the Y axis than it is to the positive part of the X axis. And the reference angle for five pi divided by three is degrees. So wanting the tangent of 60 degrees, we still want the opposite divided by the adjacent, the opposite for 60 degrees is the square of three and that's divided by the adjacent, which is one. So here this simplifies to the square of three. But very importantly, we are in the fourth quadrant is tangent positive or negative in the fourth quadrant and tangent is negative in the fourth quadrant. So we have the negative square of three for the tangent of five pi divided by three. So now we can plug both of these in to our sum identity for the tangent of A plus B. So starting off in the numerator, we have the tangent of A which is the square of three divided by three. And that's being added to the tangent of B which is the negative squarer of three. That whole thing is being divided by, in the denominator one minus the tangent of A which is the square rot of three divided by three, multiplied by the tangent of B which is the negative square of three. So many threes here. OK. So now it's just a matter of simplifying adding those in the numerator, those two terms, we want a common denominator to be able to add with fractions. So we'll multiply the negative square it of three by three divided by three. So we'll have in the numerator, the square of three divided by three. And then that one's negative. So minus three square at three divided by three or in other words, one square, it of three minus three squares of three. That's gonna be a negative two square at three divided by three in the numerator. And then we could put a division symbol. Now going to the denominator, we're multiplying the negative squared of three divided by three by the negative square it of three. So it's going to be a plus. So we'll have that one plus the product. And when we take the square, it of three multiplied by the squared of three, that's three. So this is one plus three divided by three or in other words, one plus one. So that equals two. So we have negative two square at three divided by three, divided by two. And we know that we can rewrite this. When we're dividing with fractions, we can rewrite it by multiplying by the reciprocal of the second fraction. So negative two square at three divided by three, multiplied by the reciprocal of two, which is one half. And now these twos will cancel each other out and we're essentially just multiplying by one. So this simplifies to be the negative square at three divided by three. And that is the exact value of this expression. We look at our answer choices and it matches with answer choice. D well done. We'll catch you on the next one.

My Courses

Chemistry

Biology

Math

Physics

Business

Social Sciences

Programming

Product & Marketing

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