Hey, everyone. In this problem, we're asked to find the sine of a+b given that the sine of a is equal to three fifths, the sine of b is equal to negative one-half, and we're also given that angle a is between π2 and π, and angle b is between 3π2 and 2π. So let's go ahead and get started here with our steps. Now here we're told to expand our identity and then identify any unknown trig values. Now here we're trying to find the sine of a+b. So we of course want to use our sum identity for sine. So let's go ahead and expand that out. Now the sine of a+b is going to be equal to the sine of a times the cosine of b plus the cosine of a times the sine of b. Now which of these trig values do we already know? Well, in my problem, I'm told that the sine of a is equal to three fifths and the sine of b is equal to negative one-half. So those are my two trig values that I already know. So my missing trig values are the cosine of b and the cosine of a. Now let's get started in finding those missing trig values by continuing on in our steps. Step one is done. Moving on to step number two. From all of the info that we're given in that problem statement, we want to go ahead and sketch and label our triangles. Remember that we have to pay close attention to our signs here. Let's go ahead and start with angle a. Now we're told that angle a is between π2 and π, which is right here in quadrant 2. So I want to go ahead and sketch my right triangle and label that angle a there. Now here I have my angle a. And I also know that the sine of angle a is equal to three fifths. So that tells me that my opposite side is equal to 3 and my hypotenuse is equal to 5. Now for my angle b, I'm told that angle b is between 3π2 and 2π, which is down here in quadrant 4. So we want to go ahead and sketch our triangle in quadrant 4 and then label that angle b. Now here I have my angle b and I also know that the sine of this angle is equal to negative one-half. Now that tells me that my opposite side is equal to negative one. This is a negative value because it's in the negative quadrant here for my y values. Then my hypotenuse is equal to 2, which remember is always going to be positive. So seeing as the sine value is negative one-half, that opposite side is negative one, that hypotenuse is 2. Now we have completed step number two. Moving on to step number three, we want to find any of our missing side lengths. So let's start with our triangle right here with angle b. Now here we're missing this side. So I'm going to set up my Pythagorean Theorem, a2 plus b2 equals c2. Now using my Pythagorean Theorem, I am missing one of my side lengths. So I'm going to keep a as that variable that I'm solving for. Then I'm adding that together with b2, which is negative one, and that's equal to c2, which is 2. So squaring those values gives me a2 plus 1 is equal to 4. Now if I subtract 1 on both sides, it cancels out, leaving me with a2 is equal to 3. So my final answer here is that a is equal to the square root of 3. Now you might have already noticed that that side length would be the square root of 3 because of those other two values. So if you already recognize that, that's totally fine. You don't have to go through using the Pythagorean Theorem unless you need to. Now let's look at our other triangle here. Here for my angle a, I'm missing this side length here, but my other two side lengths are 3 and 5. So this is a 3, 4, 5 triangle, and that missing side length is 4. But remember, we got to pay attention to that sign. And this is a negative 4 because of where it's located on my coordinate system. Now we've completed step number three. We can actually solve for those unknown trig values. Now our unknown trig values here, remember, were the cosine of b and the cosine of a. So let's go ahead and solve for those starting with the cosine of b. Now the cosine of b using this triangle here is going to be equal to that adjacent side, the square root of 3, over that hypotenuse of 2. Then for the cosine of a using my other triangle, the cosine of a is going to be equal to again that adjacent side negative 4 over that hypotenuse 5. Now we have all of the information that we need to actually evaluate the sum. So let's move on to step number five and finally plug in all of our values. Now doing that here, remember that the sine of a and the sine of b were already in our problem statement. We already know what they are. So the sine of a was given as three fifths. And then the cosine of b is what I just found. It is root 3 over 2. And then I'm adding that together with the cosine of a, another value that I just found, negative four-fifths, and then multiplying that by the sine of b, which was again in my problem statement as negative one-half. So now we have all of our values here, and all that's left to do is algebra. So let's go ahead and start multiplying. Now we wanna multiply these two fractions here and doing that gives me 3√3/10. Then I'm adding that together with these two fractions multiplied across. Negative four-fifths times negative one-half gives me positive 4/10. Now again, because these have a common denominator, I can go ahead and just add those across. So this is going to give me 33+410, and this is my final answer here. Please let me know if you have any questions. Thanks for watching.
Table of contents
- 0. Review of College Algebra4h 43m
- 1. Measuring Angles39m
- 2. Trigonometric Functions on Right Triangles2h 5m
- 3. Unit Circle1h 19m
- 4. Graphing Trigonometric Functions1h 19m
- 5. Inverse Trigonometric Functions and Basic Trigonometric Equations1h 41m
- 6. Trigonometric Identities and More Equations2h 34m
- 7. Non-Right Triangles1h 38m
- 8. Vectors2h 25m
- 9. Polar Equations2h 5m
- 10. Parametric Equations1h 6m
- 11. Graphing Complex Numbers1h 7m
6. Trigonometric Identities and More Equations
Sum and Difference Identities
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Sum and Difference Identities practice set
