Compute the indicated quantity using the following data. sinα=(12)/(13) where (π)/(2)<α<π

Compute the indicated quantity using the following data. ...

Compute the indicated quantity using the following data. $$\begin{aligned}&\sin \alpha=\frac{12}{13} \quad \text { where } \frac{\pi}{2}<\alpha<\pi\\&\cos \beta=-\frac{3}{5} \quad \text { where } \pi<\beta<\frac{3 \pi}{2}\\&\cos \theta=\frac{7}{25} \quad \text { where }-2 \pi<\theta<-\frac{3 \pi}{2}\end{aligned}$$ (a) $\sin (\alpha+\beta)$ (b) $\cos (\alpha+\beta)$

Compute the indicated quantity using the following data.
$$\begin{aligned}&\sin \alpha=\frac{12}{13} \quad \text { where } \frac{\pi}{2}<\alpha<\pi\\&\cos \beta=-\frac{3}{5} \quad \text { where } \pi<\beta<\frac{3 \pi}{2}\\&\cos \theta=\frac{7}{25} \quad \text { where }-2 \pi<\theta<-\frac{3 \pi}{2}\end{aligned}$$
(a) $\sin (\alpha+\beta)$
(b) $\cos (\alpha+\beta)$

Key Concepts

Quadrant and Sign Considerations

Considering the quadrant in which an angle lies is crucial because the sign of the trigonometric functions (sine and cosine) depends on the quadrant. Each quadrant of the unit circle has specific sign conventions that must be applied to the calculated values, ensuring the correct overall sign when using the angle sum formulas.

Trigonometric Angle Sum Identities

The trigonometric angle sum identities, such as sin(? + ?) = sin ? · cos ? + cos ? · sin ? and cos(? + ?) = cos ? · cos ? ? sin ? · sin ?, are essential tools for breaking down the sine and cosine of a sum of angles into expressions involving the individual angles. This simplifies the computation when certain trigonometric values of the individual angles are known.

Pythagorean Identity

The Pythagorean identity, sin²? + cos²? = 1, is fundamental in trigonometry as it allows for deriving one trigonometric function given the other. In problems like these, it facilitates finding the missing sine or cosine values from the provided data, ensuring that all components needed for the angle sum formulas are appropriately determined.

Transcript

00:01 Hello everybody.

00:02 In this video, i'm going to be showing you how to solve exercise 25 in chapter 9, section 1 of cohen's precalculus 7th edition.

00:10 In this problem, we are given two angles, alpha and beta, and we are told that the sign of alpha is equal to 1213s and lies between pi over 2 and pi, at least for the angle itself.

00:23 And as for beta, we are given that cosine of beta is equal to negative 3 5ths, and beta lies in between pi and 3 pi over 2.

00:31 And with this information, they want us to compute the sign of alpha plus beta as well as the cosine of alpha plus beta.

00:40 Now, in order to do this, we're first going to want to know the cosine of alpha as well as the sign of beta.

00:46 And to find this out, we can use a pythagorean identity, which states that for an angle t, sine squared of t plus cosine squared of t for any angle t is going to be equal to one.

01:01 Now what this gives us is that the sine of t is going to be equal to the square root of 1 minus cosine squared of t, and that the cosine of t is going to similarly be equal to 1 minus sine squared.

01:28 And so now using these identities, let's go ahead and calculate the cosine of alpha.

01:33 We have the cosine of alpha is going to be equal to the square root of 1.

01:41 Minus sine squared of alpha.

01:45 But we know that the sign of alpha is 12 .13s.

01:49 So this becomes 1 minus 12 over 13 squared.

01:58 This is all square rooted.

02:00 And upon square in that quantity, we get 144 over 169.

02:09 As we're subtracting this from 1, this becomes 169 minus 144 over 169.

02:21 This is equal to square root of 25 over 169.

02:31 And this gives us, because alpha is between pi over 2 and pi, and therefore its cosine must be negative, negative 5 over 13, which is the cosine of alpha.

02:47 And now we want to calculate the sign of beta.

02:51 Using the first of those two pythagorean identities, we know that the sign of beta is going to be equal to the square root of 1 minus cosine squared.

03:00 Of beta.

03:03 And we know that the cosine of beta is negative three -fifths.

03:07 So this is going to be equal to 1 minus negative 3 -5s squared.

03:15 Now 3 -5 squared is equal to 9 -25s.

03:22 So as we're subtracting this from 1, this becomes the square root of 25 minus 9 over 25.

03:33 25 minus 9 is 16.

03:35 So we end up with the square root of 16 over 25.

03:40 And now, because beta lies between pi and 3 pi over 2, we know that its sign must be negative.

03:47 So we're going to choose the negative square root, negative 4 fifths.

03:55 And so now we have the sign of beta as well...