Finding Exact Values Of Sin(α/2) And Tan(α/2)

Hey guys! Let's dive into a cool trigonometric problem where we're given that ${\sin \alpha = \frac{12}{13}}$ and ${90^{\circ} < \alpha < 180^{\circ}}$. Our mission, should we choose to accept it (and we do!), is to find the exact values of ${\sin \frac{\alpha}{2}}$ and ${\tan \frac{\alpha}{2}}$. Buckle up, it's going to be a fun ride!

Understanding the Problem

Before we jump into calculations, let’s make sure we understand what we're dealing with. We're given the sine of an angle ${\alpha}$, and we know that ${\alpha}$ lies in the second quadrant (between 90° and 180°). This is super important because it tells us about the signs of the other trigonometric functions of ${\alpha}$. In the second quadrant, sine is positive, cosine is negative, and tangent is negative. Keeping this in mind will help us avoid errors later on.

Our goal is to find the sine and tangent of half of this angle, ${\frac{\alpha}{2}}$. To do this, we'll need to use half-angle formulas. These formulas are our secret weapon in this mission. They relate the trigonometric functions of an angle to the trigonometric functions of half that angle. We will explore these formulas in detail in the next section. But before that, it’s crucial to visualize where ${\frac{\alpha}{2}}$ lies. If ${\alpha}$ is between 90° and 180°, then ${\frac{\alpha}{2}}$ will be between 45° and 90°. This means ${\frac{\alpha}{2}}$ is in the first quadrant, where all trigonometric functions are positive. This is another critical piece of information that will guide us in choosing the correct signs for our final answers.

So, to recap, we know the sine of ${\alpha}$, the quadrant in which ${\alpha}$ lies, and consequently, the quadrant in which ${\frac{\alpha}{2}}$ lies. We are armed with the knowledge of the signs of trigonometric functions in each quadrant. Now, let’s move on to the next step: recalling and applying the half-angle formulas. Remember, in math, just like in life, understanding the problem is half the battle. Once we have a clear picture of what we're trying to achieve, the solution becomes much more accessible. Let's get those formulas ready and solve this trigonometric puzzle!

Recalling Half-Angle Formulas

Okay, guys, let’s arm ourselves with the right tools for the job – the half-angle formulas! These formulas are essential for finding the trigonometric functions of half an angle when we know the trigonometric functions of the full angle. There are specific formulas for sine, cosine, and tangent of half an angle, and we'll use the ones for sine and tangent since that's what the problem asks for.

The half-angle formula for sine is given by: ${\sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{2}}}$

Notice the ${\pm}$ sign? That's because we need to consider the quadrant in which ${\frac{\alpha}{2}}$ lies to determine whether the sine is positive or negative. As we discussed earlier, since ${\frac{\alpha}{2}}$ is in the first quadrant, ${\sin \frac{\alpha}{2}}$ will be positive. So, we'll take the positive square root.

Now, let's look at the half-angle formula for tangent. There are a couple of ways to express this, but one common form is: ${\tan \frac{\alpha}{2} = \frac{\sin \alpha}{1 + \cos \alpha}}$

This formula is particularly handy because it directly uses ${\sin \alpha}$ and ${\cos \alpha}$, which are values we either know or can easily find. There's another form involving ${1 - \cos \alpha}$ in the denominator, but we'll stick with this one for simplicity.

So, to recap, we've got our formulas ready: the half-angle formula for sine and the half-angle formula for tangent. But wait, there's a little catch! We know ${\sin \alpha}$, but we don't yet know ${\cos \alpha}$. Bummer! But don't worry, we've got a trigonometric identity up our sleeves to help us find it. Remember the Pythagorean identity? It's a classic for a reason, and it'll come to our rescue here. Before we can plug and chug into our half-angle formulas, we need to find ${\cos \alpha}$. Let’s dive into that next!

Finding cos α

Alright, team, we've got our half-angle formulas ready to roll, but there's a missing piece in our puzzle: the value of ${\cos \alpha}$. We know ${\sin \alpha = \frac{12}{13}}$, and we know that ${\alpha}$ is in the second quadrant. This is where the fundamental trigonometric identity, the Pythagorean identity, comes to our rescue. This identity states: ${\sin^2 \alpha + \cos^2 \alpha = 1}$

This equation is like a Swiss Army knife for trigonometric problems; it's incredibly versatile. We can use it to find ${\cos \alpha}$ since we already know ${\sin \alpha}$. Let's plug in the value of ${\sin \alpha}$: ${\left(\frac{12}{13}\right)^2 + \cos^2 \alpha = 1}$

Now, let's simplify and solve for ${\cos^2 \alpha}$: ${\frac{144}{169} + \cos^2 \alpha = 1}$ ${\cos^2 \alpha = 1 - \frac{144}{169}}$ ${\cos^2 \alpha = \frac{169 - 144}{169}}$ ${\cos^2 \alpha = \frac{25}{169}}$

Now, we take the square root of both sides to find ${\cos \alpha}$: ${\cos \alpha = \pm \sqrt{\frac{25}{169}}}$ ${\cos \alpha = \pm \frac{5}{13}}$

Here's where our knowledge of the quadrant comes in clutch. Remember, ${\alpha}$ is in the second quadrant, where cosine is negative. Therefore, we choose the negative value: ${\cos \alpha = -\frac{5}{13}}$

Woohoo! We've found ${\cos \alpha}$. This was a crucial step because we need this value for both the half-angle formulas for sine and tangent. Now that we have both ${\sin \alpha}$ and ${\cos \alpha}$, we're fully equipped to tackle the half-angle formulas. We're in the home stretch now, guys. Let’s plug these values into our formulas and find the exact values of ${\sin \frac{\alpha}{2}}$ and ${\tan \frac{\alpha}{2}}$.

Calculating sin(α/2)

Okay, champions, we've found ${\cos \alpha}$, and we have the half-angle formula for sine ready to go. It’s time to put everything together and calculate ${\sin \frac{\alpha}{2}}$. Remember the formula: ${\sin \frac{\alpha}{2} = \pm \sqrt{\frac{1 - \cos \alpha}{2}}}$

We already established that ${\frac{\alpha}{2}}$ lies in the first quadrant, so ${\sin \frac{\alpha}{2}}$ is positive. We'll use the positive square root. Now, let’s plug in the value of ${\cos \alpha = -\frac{5}{13}}$: ${\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \left(-\frac{5}{13}\right)}{2}}}$

Simplify the expression inside the square root: ${\sin \frac{\alpha}{2} = \sqrt{\frac{1 + \frac{5}{13}}{2}}}$ ${\sin \frac{\alpha}{2} = \sqrt{\frac{\frac{13}{13} + \frac{5}{13}}{2}}}$ ${\sin \frac{\alpha}{2} = \sqrt{\frac{\frac{18}{13}}{2}}}$

Now, divide by 2, which is the same as multiplying by ${\frac{1}{2}}$: ${\sin \frac{\alpha}{2} = \sqrt{\frac{18}{13} \cdot \frac{1}{2}}}$ ${\sin \frac{\alpha}{2} = \sqrt{\frac{9}{13}}}$

Finally, simplify the square root: ${\sin \frac{\alpha}{2} = \frac{\sqrt{9}}{\sqrt{13}}}$ ${\sin \frac{\alpha}{2} = \frac{3}{\sqrt{13}}}$

To rationalize the denominator, we multiply both the numerator and denominator by ${\sqrt{13}}$: ${\sin \frac{\alpha}{2} = \frac{3}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}}}$ ${\sin \frac{\alpha}{2} = \frac{3\sqrt{13}}{13}}$

Boom! We've found the exact value of ${\sin \frac{\alpha}{2}}$. It might have looked a little daunting at first, but we broke it down step by step and nailed it. Now, there's just one piece of the puzzle left: finding ${\tan \frac{\alpha}{2}}$. We're on a roll, so let’s keep the momentum going and finish strong!

Calculating tan(α/2)

Alright, let's wrap this up by finding the exact value of ${\tan \frac{\alpha}{2}}$. We've already done the heavy lifting by finding ${\sin \alpha}$ and ${\cos \alpha}$, and we even calculated ${\sin \frac{\alpha}{2}}$. Now, we’ll use the half-angle formula for tangent that we discussed earlier: ${\tan \frac{\alpha}{2} = \frac{\sin \alpha}{1 + \cos \alpha}}$

This formula is super convenient because it directly uses ${\sin \alpha}$ and ${\cos \alpha}$, which we already know. Let's plug in the values we have: ${\sin \alpha = \frac{12}{13}}$ and ${\cos \alpha = -\frac{5}{13}}$: ${\tan \frac{\alpha}{2} = \frac{\frac{12}{13}}{1 + \left(-\frac{5}{13}\right)}}$

Now, let's simplify the denominator: ${\tan \frac{\alpha}{2} = \frac{\frac{12}{13}}{1 - \frac{5}{13}}}$ ${\tan \frac{\alpha}{2} = \frac{\frac{12}{13}}{\frac{13}{13} - \frac{5}{13}}}$ ${\tan \frac{\alpha}{2} = \frac{\frac{12}{13}}{\frac{8}{13}}}$

To divide fractions, we multiply by the reciprocal of the denominator: ${\tan \frac{\alpha}{2} = \frac{12}{13} \cdot \frac{13}{8}}$

Notice that the 13s cancel out: ${\tan \frac{\alpha}{2} = \frac{12}{8}}$

Now, we simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 4: ${\tan \frac{\alpha}{2} = \frac{12 \div 4}{8 \div 4}}$ ${\tan \frac{\alpha}{2} = \frac{3}{2}}$

And there we have it! The exact value of ${\tan \frac{\alpha}{2}}$ is ${\frac{3}{2}}$. We’ve successfully navigated through this trigonometric problem, using half-angle formulas and our knowledge of quadrants to guide us. Give yourselves a pat on the back, guys – we crushed it!

Conclusion

Fantastic job, everyone! We’ve successfully found the exact values of ${\sin \frac{\alpha}{2}}$ and ${\tan \frac{\alpha}{2}}$ given that ${\sin \alpha = \frac{12}{13}}$ and ${90^{\circ} < \alpha < 180^{\circ}}$. We started by understanding the problem and identifying the key information: the value of ${\sin \alpha}$ and the quadrant in which ${\alpha}$ lies.

We then recalled the half-angle formulas for sine and tangent, which were crucial for solving this problem. We realized that we needed to find ${\cos \alpha}$ before we could use the half-angle formulas, so we employed the Pythagorean identity, ${\sin^2 \alpha + \cos^2 \alpha = 1}$, to find that ${\cos \alpha = -\frac{5}{13}}$. Remember, the negative sign was important because ${\alpha}$ is in the second quadrant.

Next, we plugged the value of ${\cos \alpha}$ into the half-angle formula for sine and found ${\sin \frac{\alpha}{2} = \frac{3\sqrt{13}}{13}}$. We carefully simplified the expression and rationalized the denominator to arrive at our final answer.

Finally, we used the half-angle formula for tangent and the values of ${\sin \alpha}$ and ${\cos \alpha}$ to find ${\tan \frac{\alpha}{2} = \frac{3}{2}}$. We simplified the fraction to get our final answer.

This problem highlights the importance of understanding trigonometric identities, half-angle formulas, and the properties of trigonometric functions in different quadrants. By breaking the problem down into smaller, manageable steps, we were able to solve it effectively. So, the next time you encounter a trigonometric challenge, remember the tools and techniques we used here, and you'll be well-equipped to tackle it head-on. Keep practicing, keep exploring, and most importantly, keep having fun with math!