Given cos(A) = frac3537 (Quadrant 4) and sin(B) = frac817...

Given $cos(A) = frac{35}{37}$ (Quadrant 4) and $sin(B) = frac{8}{17}$ (Quadrant 2), a) Did you draw your reference triangles on your work page and find the missing information? (Type Yes in the box) **For parts (b) - (e), round your answers to 3 decimal places. Make sure to show all work and provide exact answers on your paper** b) Find $sin(A - B)$ c) Find $cos(2A)$ d) Find $tan(2B)$ e) Find $sin(frac{A}{2})$ **This will be worth 5 problems, so don't think it's a waste of time!** > Next Question

Transcript

00:01 So in this problem, we're given that the cosine of angle a is 35 over 7, which is a quadrant 4 angle, and that the sine of angle b is 8 over 17, which is a quadrant 1 angle.

00:10 So in part a, they want you to know if you drew your reference triangles and found out any of the missing information.

00:15 So let's start with that.

00:16 So first, let's graph or graph angle a.

00:19 Well, we know it's a 4th quadrant angle, so it's going to look like this.

00:22 So this would be angle a.

00:24 And we know the cosine of angle a is 35 over 7.

00:27 Remember, cosine is the ratio of the adjacent side over the hypotenuse.

00:32 So we just need to solve for that opposite side by using the pythagorean theorem.

00:36 So we're going to have x squared plus 35 squared equal to 37 squared.

00:41 Well, 35 squared is equal to 1 ,225.

00:47 37 squared is equal to 1 ,369.

00:52 So to solve for x, we're going to subtract 1 ,225 from both sides of this equation.

00:58 So if we take 1 ,369 and subtract 1 ,225, that's equal to 144.

01:05 And then we'll take the square root of both sides, and the square root of 144 is equal to 12.

01:10 So now we can go back and label our triangle.

01:13 Keep in mind, because it's a 4th quadrant, that height will be a negative value.

01:17 So it'll be negative 12.

01:20 So now we have to do the same thing for angle b.

01:22 Now, it's a 1st quadrant angle, so we'll draw it in the 1st quadrant.

01:26 Here's angle b.

01:27 It's the sine, which is the ratio of the opposite side over the hypotenuse.

01:31 So we can find that adjacent side, again, by using pythagorean theorem.

01:35 So we'll have y squared plus 8 squared equal to 17 squared.

01:39 8 squared is equal to, not equals there, 8 squared is equal to 64.

01:44 And we have 17 squared, which is equal to 289.

01:48 Then i'm going to subtract 64 from both sides of our equation.

01:52 So we have y squared equal to, well, 289 minus 64 is 225.

01:58 So when we take the square root of both sides, the square root of 225 is 15.

02:02 So because we're in the 1st quadrant, all of our sides will be positive.

02:05 So now we know that y, or that adjacent side, is 15.

02:09 So perfect.

02:10 Part a is now all set.

02:12 So now we can take a look at the next part, part b.

02:16 Part b says to find the sine of angle a minus angle b.

02:20 So to do this, we have to use our difference formula for sine, which says that the sine of angle a times the cosine of angle b minus the cosine of angle a times the sine of angle b.

02:33 So we can go ahead and use our triangles to help us.

02:37 So the sine of angle a, that's the opposite side over the hypotenuse, so negative 12 over 37, times the cosine of angle b, that's the adjacent side over hypotenuse, so 15 over 17, minus the cosine of angle a, which we were told was 35 over 37, times the sine of angle b, which we were given was 8 over 17...

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