Use the following conditions to solve Exercises 1–4:4 𝝅sin α = -...

Question

Use the following conditions to solve Exercises 1–4:4 𝝅sin α = ----- , ------- < α < 𝝅5 25 𝝅cos β = ------ , 0 < β < ------13 2Find the exact value of each of the following.cos (α + β)

Accepted Answer

Hello, today we're gonna be determining the exact value of the given expression. So what we want to find is cosine of A minus B. Now, before we find the value of this expression, we can first expand this expression by using the difference identity for cosine the difference identity for cosine states that we can expand cosine A minus B as cosine of A multiplied by cosine of B plus sign of A multiplied by sign of B. So in order to find the exact value of cosine A minus B, we're going to need to find the values of cosine of a cosine of BS sign of A and S sign of B. And we can find these values by using the given conditions. Let's take a look at the first condition. The first condition states that sign of A is equal to 35/37. And the angle of A lies in the region between pi over two and pi. Now, one thing to note is that the region of the unit circle between Pi over two and Pi is going to be quadrant three. And within quadrant three, any X value is going to be negative and any Y value will be negative. So how can we use sine of A in order to solve for cosine of a? Well, we can go ahead and use a right triangle in the third quadrant. So let's go ahead and draw that triangle. Now recall that any sign value for any right triangle is defined as the height of the triangle over its hypotenuse and cosine of any right triangle is defined as the length of the triangle over its hypotenuse. If we use the identity for sine and the given value for sign of a, we can assign the triangle a height of 35 the hypotenuse a length of 37. Now recall that any Y value in this region is negative. So the height of the triangle is going to be negative 35. So how can we solve for the length of the triangle? Well, we can use the Pythagorean theorem to solve for the missing length. And the Pythagorean theorem is given to us as the hypotenuse squared is equal to X squared, which is the length plus the height squared. In this case, Y will it be negative 35 H will be 37. We can plug in these values to the equation to give us 37 squared is equal to X squared plus negative 35 squared, 37 squared will give us the value of 1369. A negative 35 squared will give us the value of 1225. We'll subtract 1225 to both sides of the equation. Leaving this with X squared is equal to 144. And taking the square root of both sides of this equation will give us an X value of 12. So the length of the triangle is going to be 12. But again, any X value within the third quadrant will be negative. So the length of the triangle will be negative 12. And now that we have the length of the triangle, we can find the value of cosine of A, using the identity for cosine of a right triangle, cosine of A will equal to negative 12/37. So we've solved for our cosine of A and S sign of A values. But now we need to solve for cosine of B and sign of B. Let's take a look at the second condition given to us. The second condition states that cosine of B is equal to 20/29. And the angle of B is in the region between zero and pi over two. Now, the region of the unit circle between zero and pi over two is going to be quadrant one and any values of X within quadrant one will be positive and any Y values will be positive. Now, just like before we're going to use A right triangle and cosine of B. In order to solve for sign of B, we're going to draw a right triangle in quadrant one. And using cosine of B, we can assign a length of 20 and a hypotenuse of 29. And plugging these values into the Pythagorean theorem will give us the equation 29 squared is equal to 20 squared plus Y squared. 29 squared will give us the value of 841 and 20 squared will give us the value of 400. If we subtract 400 to both sides, we get Y squared is equal to 441. And taking the square root of both sides of the equation will give us Y is equal to 21. So the height of the right triangle will be 21. So using the identity for sign sign of A is going to equal to 21/29. So now that we have our sign of a sign of B value, our cosine of A and cosine of B value, we can go ahead and plug this in to the given equation to solve for the exact value of cosine A minus B. So if we plug this in, we end up with this expression negative 12/ multiplied by 20/ plus 35/ multiplied by 21/ negative 12/37 multiplied by 20/29 is going to give us the value of negative 240 over 1073 and 35/37 multiplied by 21/29 will give us the value of 735 over 1073. And adding these two fractions together will give us a final value of 495 over 1073. So what this means that the expression cosine A minus B will have the exact value of 495 over 1073. And this is going to be the answer to the problem. And with that being said, the overall solution 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.

Use the following conditions to solve Exercises 1–4:4 𝝅sin α = ----- , ------- < α < 𝝅5 25 𝝅cos β = ------ , 0 < β < ------13 2Find the exact value of each of the following.cos (α + β)

Key Concepts

Sine and Cosine Functions

Angle Addition Formula

Quadrants and Angle Ranges

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