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:

β
e. cos -------
2

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

Accepted Answer

Hello, today we're gonna be determining the value of the given expression. So we are determining the value of cosine B over two. Now, before we start determining the value, let's go ahead and expand the given expression, we can expand the expression by using the half angle formula for cosine and the half angle for co formula for cosine states that if you have cosine data over two, this expression can be expanded as plus or minus the square root of one plus cosine theta all divided by two. So if we use this identity, we can rewrite the given expression as plus or minus the square root of one plus cosine B over two. Now, what we'll need to do in order to get the exact value of cosine B over two is we'll need to first determine the value of cosine of B. In order to do this, we can use the given condition for the angle B. Now the condition for angle B states that we have sin of B is equal to negative 1/7. And the angle of B lies in the region between pi and three pi over two. Now one thing to note is that the region of the unit circle between pi and pi three, pi over two is going to be quadrant three in quadrant three, any cosine value is going to be negative and any sign value will be negative. Furthermore, any X value in this region will be negative and any Y value in this region will be negative. So how do we solve for cosine and B if we are given sign, well, we can go ahead and use a right triangle that exists in quadrant three. Now recall that the identity for sign of any right triangle is given to us as the height of the triangle over its hypotenuse. And the identity for cosine of any right triangle will be the length of the triangle over its hypotenuse. So using the definition of sine, we can re we can label the side links of the triangle as a height of one and a hypotenuse of seven. Now keep in mind any Y value in this region is going to be negative. So the triangle is going to have a height of negative one. So what we need is we need the length of the triangle. In order to get cosine, we can solve for the length of the triangle by using the Pythagorean theorem. And the Pythagorean theorem states that the hypotenuse 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 plugging this into the equation will give a seven squared is equal to X squared plus negative one squared, seven squared will evaluate to give us 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 which will give us X squared is equal to 48. And taking the square root of both sides of this equation will give us X is equal to four square roots of three. So the length of this triangle in quadrant three will be four square root of three. But again, any X value within this region will be negative. So the triangle is going to have a length of negative four square root of three. Now that we have the X value and we have the cosine or the hypotenuse, we can use the identity of cosine to solve for cosine of B and cosine of B. In this case is going to be negative four square root of 3/7. Now that we have the value for cosine of B, we can plug this into the half angle identity to solve for cosine B over two. So if we plug this in to the identity, we get plus or minus the square root of one minus four square root of 3/7, all divided by two, let's start by simplifying the inside numerator. So the numerator is one minus four square root of 3/7, we can rewrite one at 7/7 and we can have the numerator at 7/7 minus four square root of 3/7. And if we take the difference of these two fractions, we get a numerator of seven minus four square root of 3/7. So we can rewrite this expression as plus or minus the square root of seven minus four square root of 3/7, all over two. So what we are left with is a complex fraction inside the square root. We can express the denominator as a fraction by putting it over one. And to simplify the complex fraction, we multiply the denominator by its reciprocal and multiply the numerator by the reciprocal of the denominator. The denominator was simplified to one leaving us with seven minus four square root of 3/7, multiplied by 1/2. And multiplying these values together is going to give us the expression of plus or minus the square root seven minus four square root of 3/14. So this is the simplified form of cosine B over two. But keep in mind that this is two solutions, we have a positive version and a negative version of the solution. So which version do we pick? Well, the quadrant answers that question again, the angle of B lies within quadrant three of the unit circle and any cosine value within quadrant three is going to be negative. So we're going to be picking the negative version of this solution. And that means cosine of B over two will equal to the negative square root of seven minus four square root of three, all over 14. This is going to be the answer to this problem. And with that being said, the overall answer is going to be D. 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: β e. cos ------- 2 1 3𝝅 1 3𝝅 sin α =﹣ ------ , 𝝅 < α < ------- , and cos β =﹣------ , 𝝅 < β < ---------. 3 2 3 2

Key Concepts

Trigonometric Functions

Reference Angles

Quadrant Analysis

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