Find sin(beta - alpha) given sin(alpha) = 15/17 and tan(beta) = 12/5 .

Trigonometric Problem: Find \(\sin(\beta - \alpha)\)

We need to determine the value of \(\sin(\beta - \alpha)\) given the following information:

• \(\frac{\pi}{2} < \alpha < \pi\) ( \(\alpha\) is in the second quadrant)
• \(\pi < \beta < \frac{3\pi}{2}\) ( \(\beta\) is in the third quadrant)
• \(\sin \alpha = \frac{15}{17}\)
• \(\tan \beta = \frac{12}{5}\)

Options:

a) \(-\frac{171}{221}\) b) \(-\frac{21}{221}\) c) \(\frac{21}{221}\) d) \(\frac{171}{221}\)

Detailed Solution

Hello, student! Let’s solve this trigonometric problem step by step to find \(\sin(\beta - \alpha)\). We’ll use a key formula and calculate everything we need based on the information given. Don’t worry—I’ll explain everything clearly!

Step 1: Understand the Formula

To find \(\sin(\beta - \alpha)\), we use the **sine difference formula**:

\[ \sin(\beta - \alpha) = \sin \beta \cos \alpha - \cos \beta \sin \alpha \]

This formula tells us we need four values: \(\sin \beta\), \(\cos \alpha\), \(\cos \beta\), and \(\sin \alpha\). we’re already given \(\sin \alpha = \frac{15}{17}\). Let’s find the rest!

Step 2: Find \(\cos \alpha\)

Since \(\alpha\) is in the second quadrant (\(\frac{\pi}{2} < \alpha < \pi\)), sine is positive, but cosine is negative. We can use the Pythagorean identity

\[ \sin^2 \alpha + \cos^2 \alpha = 1 \]

Plug in \(\sin \alpha = \frac{15}{17}\):

\[ \left(\frac{15}{17}\right)^2 + \cos^2 \alpha = 1 \] \[ \frac{225}{289} + \cos^2 \alpha = 1 \] \[ \cos^2 \alpha = 1 - \frac{225}{289} = \frac{289 - 225}{289} = \frac{64}{289} \] \[ \cos \alpha = \pm \sqrt{\frac{64}{289}} = \pm \frac{8}{17} \]

Because \(\alpha\) is in the second quadrant (where cosine is negative), we take the negative value:

\[ \cos \alpha = -\frac{8}{17} \] Step 3: Find \(\sin \beta\) and \(\cos \beta\)

We’re given \(\tan \beta = \frac{12}{5}\), and \(\beta\) is in the third quadrant (\(\pi < \beta < \frac{3\pi}{2}\)), where both sine and cosine are negative. The tangent is defined as:

\[ \tan \beta = \frac{\sin \beta}{\cos \beta} \]

Think of a right triangle: if the opposite side is 12 and the adjacent side is 5, we can find the hypotenuse using the Pythagorean theorem:

\[ \text{Hypotenuse} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]

For the reference angle in this triangle:

\[ \sin(\text{reference angle}) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{13} \] \[ \cos(\text{reference angle}) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{13} \]

Since \(\beta\) is in the third quadrant (where sine and cosine are negative), we adjust the signs:

\[ \sin \beta = -\frac{12}{13} \] \[ \cos \beta = -\frac{5}{13} \]

Let’s check: \(\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{-\frac{12}{13}}{-\frac{5}{13}} = \frac{12}{5}\), which matches what we’re given. Perfect!

Step 4: Compute \(\sin(\beta - \alpha)\)

Now we have all the pieces:

• \(\sin \alpha = \frac{15}{17}\)
• \(\cos \alpha = -\frac{8}{17}\)
• \(\sin \beta = -\frac{12}{13}\)
• \(\cos \beta = -\frac{5}{13}\)

Substitute into the sine difference formula:

\[ \sin(\beta - \alpha) = \sin \beta \cos \alpha - \cos \beta \sin \alpha \]

First term: \(\sin \beta \cos \alpha\)

\[ \left(-\frac{12}{13}\right) \left(-\frac{8}{17}\right) = \frac{12}{13} \cdot \frac{8}{17} = \frac{12 \cdot 8}{13 \cdot 17} = \frac{96}{221} \]

(A negative times a negative gives a positive result.)

Second term: \(\cos \beta \sin \alpha\)

\[ \left(-\frac{5}{13}\right) \left(\frac{15}{17}\right) = -\frac{5 \cdot 15}{13 \cdot 17} = -\frac{75}{221} \]

Now combine them:

\[ \sin(\beta - \alpha) = \frac{96}{221} - \left(-\frac{75}{221}\right) = \frac{96}{221} + \frac{75}{221} = \frac{96 + 75}{221} = \frac{171}{221} \] Step 5: Verify the Sign

Let’s make sure this makes sense. Since \(\pi < \beta < \frac{3\pi}{2}\) and \(\frac{\pi}{2} < \alpha < \pi\), \(\beta\) is greater than \(\alpha\). So:

\[ \beta - \alpha > 0 \]

The maximum value of \(\beta - \alpha\) is \(\frac{3\pi}{2} - \frac{\pi}{2} = \pi\), and the minimum is positive. Thus:

\[ 0 < \beta - \alpha < \pi \]

This means \(\beta - \alpha\) is in the first or second quadrant, where sine is positive. Our result, \(\frac{171}{221}\), is positive, so it checks out!

Final Answer

The value of \(\sin(\beta - \alpha) = \frac{171}{221}\), which matches option d.