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

3 𝝅 12 𝝅 
sin α = ------- , 0 < α < -------- , and sin β = --------- , --------- < β < 𝝅. 
5 2 13 2

Accepted Answer

Hello, today we're gonna be finding the exact value of the given expression. So what we are given is cosine of A over two. Now, in order to find the exact value of this, we can first expand cosine A over two by using the half angle identity for cosine the half angle identity for cosine states that we can expand the given expression as plus or minus the square root of one plus cosine of A that quantity divided by two. So in order to get the exact value of cosine A over two, we're first going to solve for the value of cosine of A. And we can do that by using the given condition for the angle of a, the condition states that sign of A is equal to 7/25. And the angle of A exists in the region of the unit circle between zero and pi over two. Now, one thing to note is that the region of the unit circle between zero and pi over two is going to be quadrant one within this quadrant, any cosine value and sine value is going to be positive. Furthermore any X value or Y value will be positive. So how can we use sine of a in order to solve for cosine of a? Well, we can go ahead and use the right triangle that exists in quadrant one. Now recall that the identity for any sign value of a right triangle is defined as the height of the triangle over its hypotenuse. And the cosine value for any right triangle is defined as the length of the triangle over its hypotenuse. So using the identity for the sine value, we can label the side links of the triangle as a height of seven and a hypo of 25. But now, what we'll need to do is we'll need to solve for the length. In order to get our cosine value, we can solve the length by using the hype. The Pythagorean theorem. The Pythagorean theorem states that the hypotenuse squared is equal to the length squared plus the height squared. In this case, Y will be seven and the hypotenuse will be 25. And plugging this into the Pythagorean theorem will give us 25 squared is equal to our length squared plus seven squared, 25 squared will give us the value of and seven squared will give us the value of 49. In order to solve for X, we subtract 49 to both sides of the equation and that leaves us with X squared is equal to 576 and taking the square root of both sides of this equation is going to give us X is equal to 24. So the triangle in quadrant one is going to have a value of 24 for its length. And now that we have the side length, we can go ahead and use the identity for cosine in order to solve for cosine of a, using this triangle cosine of A is going to equal to 24/25. Now that we have the value of cosine of A, we can plug in this value to the half angle identity in order to solve for cosine A over two. If we plug in this value, we get plus or minus the square root of one plus 24/25 divided by two. Let's start by simplifying the numerator in the numerator. We have one plus 24/25. We can rewrite one as 25/25 the re numerator is going to be 25/25 plus 24/25 and adding these two fractions together is going to give us the value of 49/25. So the numerator is going to simplify to 49/25 we can rewrite the expression as plus or minus the square root of 49/25 all over two. Now, what we need to do is we need to simplify the complex fraction, we can express the denominator by putting 2/1. And in order 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 49/ multiplied by 1/2 in the numerator. And that's going to simplify the expression to plus or minus the square root of 49. Over 50 we can separate the square roots in both the numerator and denominator and rewrite the expression as plus or minus the square root of 49 over the square root of the square root of 49 will simplify to give us seven. So we have plus or minus seven over the square root of we cannot leave a square root in the denominator. So we'll need to rationalize this quantity. So we'll multiply the numerator and denominator by the square root of 50. And that is going to leave us with plus or minus seven multiplied by the square root of 50/50. Now the square root of 50 is going to simplify to give us five square root of two. So in the numerator, we have seven multiplied by five, multiplied by the square root of two, all over 50. And since we have multiples of five in both the numerator and denominator, we can simplify 5 to 1 in the numerator and simplify 5 50 to 10 in the denominator that is going to leave us with the value of plus or minus seven square root of 2/10. This is going to be the simplified form of the solution. But keep in mind, we have two solutions. Currently, we have a positive version and a negative version of the solution. So the question is which version do we pick? Well, that depends on the quadrant that the angle is located in. Recall that the angle of A is located in quadrant one and any cosine value in quadrant one is going to be positive. So we're going to pick the positive version of the solution. And what that means is that cosine of A over two is going to have the exact value of seven square root of 2/10. And this is going to be the solution to this 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.

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In Exercises 35–38, find the exact value of the following under the given conditions: β e. cos ------- 2 3 𝝅 12 𝝅 sin α = ------- , 0 < α < -------- , and sin β = --------- , --------- < β < 𝝅. 5 2 13 2