In Exercises 57–64, find the exact value of the following under t...

Question

In Exercises 57–64, find the exact value of the following under the given conditions:

a. cos (α + β)

3 1
tan α = ﹣ ------ , α lies in quadrant II, and cos β = ------- , β lies in quadrant I.
4 3

Accepted Answer

Hello, today we're gonna be determining the exact value of cosine of X plus Y using the given conditions. Now, in order to determine the value of cosine of X plus Y, we first need to expand this by using the sum property of cosine, the sum property of cosine states that we can expand cosine of X plus Y as cosine of X multiplied by cosine of Y minus sine of X multiplied by sine of Y. So what we'll need to do is we'll need to solve for a cosine of X cosine of Y sign of X and sine of Y values. So how do we go about doing that? While we need to use the given conditions in order to get each of the respective angles. Let's take a look at the first condition in order to solve for cosine of X and sin of X. The first condition states that tangent of X is equal to negative 9/40. And the angle of X lies in the second quadrant of the unit circle since this angle exists in the second quadrant of the unit circle, any X value is going to be negative. And any Y value is going to be positive. So how do we use tangent of X to solve for cosine of X and sin of X? Well, we can go ahead and use right triangles. So let's create a right triangle that exists in quadrant two. Now recall that tangent of X for any right triangle is defined as the height of the triangle over the length of the triangle. In this case here, the height of the triangle is going to be nine and the length is going to be 40. But recall that any X value or the length of the triangle is negative in the second quadrant and any height of the triangle or Y value is positive. So that means that the length of the triangle is going to be negative 40. Now, we need to use this right triangle to get our cosine and sine value. Recall that cosine of X for any right triangle is defined as the length of the triangle over its hypo news and sine is defined as the height of the triangle over its hypotenuse. Currently, we are missing the hypotenuse of this right triangle, but we can solve for the hypotenuse by using the Pythagorean theorem. The Pythagorean theorem states that the hypotenuse squared is equal to the length squared plus the height squared X is going to be negative 40 Y is going to be nine. So plugging this into the equation is going to give us hypotenuse squared is equal to nine squared plus negative 40 squared, nine squared is going to evaluate to give us 81. A negative 40 squared is going to evaluate to give us 1600 and adding these values together is going to give us hypotenuse squared is equal to 1681. And taking the square root of both sides of this equation is going to give us a hypotenuse of 41. So we know that the length of the hypotenuse is going to be 41. Now that we have the hypo news, we can solve for cosine of X and sin of X cosine of X is going to equal to negative 40/41 and sign of X is going to equal to positive 9/41. Now we have our cosine of X and our sign of X value, but we still need to solve for our cosine of Y and S sign of Y values. In order to do that, let's take a look at the second condition. The second condition states that cosine of Y is equal to 1/5. So we are given the cosine of Y value. We are also told that the angle of Y lies in quadrant one of the unit circle. And since this angle lies in quadrant one, any X and Y values are going to be positive in this region. So let's solve first sign of Y by creating a secondary right triangle. And we're gonna make a right triangle that exists in quadrant one. Now again, cosine is defined as the length of the triangle over its hypotenuse. So we can use cosine of Y equal to 1/5 to give the length of the triangle one and the Typo News five. So what we'll need to do is we'll need to solve for the height of the right triangle in order to get sign of Y again, we can use the Pythagorean theorem. But this time we're going to be solving for Y since we need the height of the triangle H is going to be five and X is going to be one. So plugging this into the Hypo or into the Pythagorean theorem is going to give us five squared is equal to one squared plus Y squared. Five squared, evaluates the 25 and one squared is going to give us one subtracting one to both sides of this equation is going to give us 24 is equal to Y squared. And taking the square root of both sides of this equation is going to give us Y is equal to two square root of six. So the height of this triangle is going to be two square root of six. Now that we have at the height of the triangle we can solve for sign of Y and sin of Y is going to equal to two square root of 6/5. So just to recap, we use the first condition to solve for cosine of XS sign of X. We were given cosine of Y and used that to solve for sign of Y. Now, what we need to do is we need to take these cosine and sine values and plug it in to the sum identity of cosine in order to solve for cosine of X plus Y. So we have cosine of X which we saw for is negative 40/41 multiplied by cosine of Y which is 1/5. And we're subtracting sine of X which we solved as 9/41 multiplied by sine of Y which is two square root of 6/5. Now, what we need to do is we need to simplify negative 40/41 multiplied by 1/5 is going to give us the value of negative 40/205 9/41 multiplied by two square root of 6/5 is going to simplify to give us 18 square root of six over 205. If we subtract both of the fractions together, we can, we can simplify the fractions as negative 40 minus 18 square root of 6/205. And if we factor out a negative one and reorder the terms in the numerator, we can rewrite the fraction as negative one multiplied by the quantity 18 squared of six plus 40/205. And this is going to be the exact value for cosine of X plus Y. And with that being said, the answer to this problem is going to be 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 57–64, find the exact value of the following under the given conditions: a. cos (α + β) 3 1 tan α = ﹣ ------ , α lies in quadrant II, and cos β = ------- , β lies in quadrant I. 4 3

Key Concepts

Trigonometric Identities

Quadrants and Angle Properties

Finding Trigonometric Values from Tangent and Cosine

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