In Exercises 35–38, find the exact value of the following under t...

Question

In Exercises 35–38, find the exact value of the following under the given conditions:

b. cos(α﹣β)

1 3𝝅 1 3𝝅 
sin α =﹣ ------ , 𝝅 < α < ------- , and cos β =﹣------ , 𝝅 < β < ---------. 
3 2 3 2

Accepted Answer

Hello, today we're going to be solving for the following expression. So what we want to solve is cosine A minus B. Now, before we start solving the value of cosine A minus B, we can expand this expression by using the difference identity of cosine the difference identity for cosine states that cosine A minus B can be expanded as cosine of A multiplied by cosine of B plus sign of A multiplied by sign of B. So what we'll need to do is we'll need to solve for cosine of a cosine of B sign of A and sign of B. In order to get the exact value of cosine A minus B, we can solve for these values by using the given conditions. So let's take a look at the first condition. The first condition states that sign of A is equal to negative 1/7. Furthermore, the angle A exists in the region of the unit circle between pi and three pi over two. Now, one thing to note about this condition is that the region of the unit circle between pi and three pi over two is going to be quadrant three of the unit circle quadrant three of the unit circle has negative X values and negative Y values. So how do we use sine of A to solve for cosine of a? Well, we can use a right triangle that exists in the third quadrant. So the triangle is going to have this shape. Now recall that the identity for sine for any right triangle is defined as the height of the triangle over its hypotenuse. Furthermore, cosine of any right triangle is defined as the length of the triangle over the hypotenuse. So using the identity for sign, we can use the given sign value to label the side links of the triangle and the height of the triangle will be one and the hypotenuse will be seven. Now recall that N Y value in quadrant three is going to be negative. So the height of the triangle will be negative one. And in order to get cosine of a, we need to solve for the length of the triangle. In order to do that, we can use the Pythagorean theorem. And the Pythagorean theorem states that the hypotony squared is equal to the length squared plus the height squared. In this case, Y will be negative one and H will be seven. So we get seven squared is equal to X squared plus negative one squared. Seven squared will give us the value of 49 negative one squared will give us the value of positive one. In order to solve for X, we subtract one of both sides of the equation to give us X squared is equal to 48. And taking the square root of both sides of the equation will give us X is equal to four square root of three. So the length of the triangle at quadrant three will be four square root of three. But again, any X value within quadrant three will be negative. So the length of this triangle is going to be negative four square root of three. And now, if we use the identity of cosine for a right triangle, we get cosine of A is equal to negative four square root of three over seven. So we have our cosine of a value and we have our sign of a value we need to solve for cosine of B and sign of B. And in order to do that, we can take a look at the second condition given to us. Now, the second condition states that sign of B is equal to negative 1/7 and the angle of B exists in the region of the unit circle between pi and three pi over two. Now, because this region or the angle of B exists within the same region as the angle of A. In order to solve for cosine, we're going to use the same triangle as before we're going to have a length of four square root of three, we're going to have a height of negative one and we'll have a hypo of seven and using the identity for cosine cosine of B will equal to negative four square root of 3/7. So we have our cosine of B value and we have our sign of B value. Now that we have the four values needed, we can solve our cosine of A minus B by using the different property of cosine. So plugging these values in cosine of A is going to be negative four square root of three over seven multiplied by cosine of B which is negative four square root of 3/ plus sine of A, which is given to us as negative 1/7 multiplied by sign of B which is given to us as negative 1/7. Now let's take the product of these values. We'll start by multiplying negative four square of 3/7 with negative four square of 3/7, multiplying two negative numbers together will give us a positive number and we'll multiply the coefficients of the numerators forward first and four times four will give us next square root of three multiplied by square of three will simplify to give us three and seven times seven in the denominator will give us the value of 49. Now we'll multiply the values of negative 1/7 with negative 1/7. Again, multiplying two negative numbers gives us a positive number one times one will give us one. And in the denominator, we have seven times seven, which is 49 16 times three will give us the value of 48. So we have 48/49 plus 1/ adding these fractions together will give us the value of 49/ which is going to simplify to give us one. So what this means is that the exact value of cosine A minus B is going to equal to one and this is going to be the solution to the problem. And with that being said, the overall answer is going to be B. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

In Exercises 35–38, find the exact value of the following under the given conditions: b. cos(α﹣β) 1 3𝝅 1 3𝝅 sin α =﹣ ------ , 𝝅 < α < ------- , and cos β =﹣------ , 𝝅 < β < ---------. 3 2 3 2

Key Concepts

Trigonometric Identities

Quadrants and Angle Ranges

Exact Values of Trigonometric Functions

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