Solving Sin(x/2) Given Csc(x) = 9: A Step-by-Step Guide

Hey guys! Today, we're diving into a fun trigonometric problem where we need to find the value of $\sin(x/2)$ given that $\csc(x) = 9$ and $90^{\circ} < x < 180^{\circ}$. This problem might seem a bit tricky at first, but don't worry, we'll break it down step by step so it's super clear and easy to follow. We'll be using some key trigonometric identities and a little bit of algebraic manipulation to get to the solution. So, grab your calculators (or your mental math muscles!), and let's get started!

Understanding the Problem

In this section, we'll really nail down the heart of the problem. The core of this trigonometric problem lies in understanding the relationships between different trigonometric functions and how they behave in specific quadrants. We are given that $\csc(x) = 9$, which is the reciprocal of the sine function. This means that $\sin(x) = 1/9$. Also, we know that the angle $x$ lies in the second quadrant ($90^{\circ} < x < 180^{\circ}$). This piece of information is crucial because it tells us about the signs of the trigonometric functions in that quadrant. In the second quadrant, sine is positive, while cosine is negative. This is a super important detail that will guide us in the later steps.

Now, our ultimate goal is to find $\sin(x/2)$. To do this, we'll need to use the half-angle formula for sine. The half-angle formulas are special trigonometric identities that allow us to express trigonometric functions of half an angle in terms of trigonometric functions of the full angle. Specifically, the half-angle formula for sine is given by:

$$\sin(\frac{x}{2}) = \pm \sqrt{\frac{1 - \cos(x)}{2}}$$

Notice the "$\pm$" sign in front of the square root. This is because the sine of half an angle can be either positive or negative, depending on the quadrant in which $x/2$ lies. To determine the correct sign, we need to figure out the range of $x/2$. Since $90^{\circ} < x < 180^{\circ}$, dividing all parts of the inequality by 2 gives us $45^{\circ} < x/2 < 90^{\circ}$. This means that $x/2$ lies in the first quadrant, where sine is positive. Therefore, we'll take the positive square root.

So, to find $\sin(x/2)$, we need to find $\cos(x)$. This is where the Pythagorean identity comes in handy. The Pythagorean identity is a fundamental trigonometric identity that relates sine and cosine:

$$\sin^2(x) + \cos^2(x) = 1$$

We know $\sin(x)$, so we can use this identity to solve for $\cos(x)$. Remember that since $x$ is in the second quadrant, $\cos(x)$ will be negative. Once we have $\cos(x)$, we can plug it into the half-angle formula for sine and calculate the value of $\sin(x/2)$.

In summary, understanding the problem involves recognizing the given information, identifying the appropriate trigonometric identities (half-angle formula and Pythagorean identity), and considering the signs of trigonometric functions in different quadrants. This lays the groundwork for a clear and efficient solution.

Utilizing Trigonometric Identities

Okay, let's get our hands dirty with some actual calculations! The first crucial step involves using trigonometric identities to bridge the gap between what we know ($\csc(x) = 9$) and what we want to find ($\sin(x/2)$). As we discussed earlier, the Pythagorean identity and the half-angle formula are our best friends here. We've already established that $\sin(x) = 1/9$. Now, let's use the Pythagorean identity to find $\cos(x)$:

$$\sin^2(x) + \cos^2(x) = 1$$

Plug in $\sin(x) = 1/9$:

$$(\frac{1}{9})^2 + \cos^2(x) = 1$$

$$\frac{1}{81} + \cos^2(x) = 1$$

Now, isolate $\cos^2(x)$:

$$\cos^2(x) = 1 - \frac{1}{81} = \frac{80}{81}$$

Take the square root of both sides:

$$\cos(x) = \pm \sqrt{\frac{80}{81}} = \pm \frac{\sqrt{80}}{9} = \pm \frac{4\sqrt{5}}{9}$$

Remember, $x$ is in the second quadrant, where cosine is negative. So, we choose the negative sign:

$$\cos(x) = -\frac{4\sqrt{5}}{9}$$

Great! We've found $\cos(x)$. Now we can finally use the half-angle formula for sine:

$$\sin(\frac{x}{2}) = \pm \sqrt{\frac{1 - \cos(x)}{2}}$$

We already determined that $x/2$ is in the first quadrant, where sine is positive, so we'll use the positive square root. Plug in our value for $\cos(x)$:

$$\sin(\frac{x}{2}) = \sqrt{\frac{1 - (-\frac{4\sqrt{5}}{9})}{2}}$$

$$\sin(\frac{x}{2}) = \sqrt{\frac{1 + \frac{4\sqrt{5}}{9}}{2}}$$

To simplify this, let's find a common denominator in the numerator:

$$\sin(\frac{x}{2}) = \sqrt{\frac{\frac{9 + 4\sqrt{5}}{9}}{2}}$$

Now, divide by 2 (which is the same as multiplying by 1/2):

$$\sin(\frac{x}{2}) = \sqrt{\frac{9 + 4\sqrt{5}}{18}}$$

This is a perfectly valid answer, but we can simplify it further to get it into a cleaner form. To do this, we can try to express the numerator as a perfect square. This is a bit of a trick, but it's a useful technique to know. We're looking for numbers $a$ and $b$ such that:

$$(a + b)^2 = a^2 + 2ab + b^2 = 9 + 4\sqrt{5}$$

If we let $a^2 + b^2 = 9$ and $2ab = 4\sqrt{5}$, then $ab = 2\sqrt{5}$. By trial and error (or a bit of clever guessing!), we can find that $a = 2$ and $b = \sqrt{5}$ works:

$$(2 + \sqrt{5})^2 = 4 + 4\sqrt{5} + 5 = 9 + 4\sqrt{5}$$

So, we can rewrite our expression as:

$$\sin(\frac{x}{2}) = \sqrt{\frac{(2 + \sqrt{5})^2}{18}}$$

Now, take the square root:

$$\sin(\frac{x}{2}) = \frac{2 + \sqrt{5}}{3\sqrt{2}}$$

Finally, let's rationalize the denominator by multiplying the numerator and denominator by $\sqrt{2}$:

$$\sin(\frac{x}{2}) = \frac{(2 + \sqrt{5})\sqrt{2}}{3\sqrt{2}\sqrt{2}} = \frac{2\sqrt{2} + \sqrt{10}}{6}$$

So, our final simplified answer is:

$$\sin(\frac{x}{2}) = \frac{2\sqrt{2} + \sqrt{10}}{6}$$

Woohoo! We made it! We successfully used the Pythagorean identity and the half-angle formula, along with some algebraic manipulation, to find the value of $\sin(x/2)$.

Simplifying the Solution

Now that we've got our answer, let's talk about simplifying the solution to its most elegant form. This is a crucial skill in mathematics, guys, because it not only makes your answer look cleaner but also helps in further calculations or comparisons if needed. We arrived at $\sin(\frac{x}{2}) = \sqrt{\frac{9 + 4\sqrt{5}}{18}}$, which is correct but not the most simplified form.

As we saw in the previous section, the key to simplifying this further lies in recognizing that the expression inside the square root can be rewritten as a perfect square. Specifically, we identified that $9 + 4\sqrt{5}$ can be expressed as $(2 + \sqrt{5})^2$. This step is not always obvious, but with practice, you'll start recognizing these patterns. One clue is the presence of a square root term (like $4\sqrt{5}$), which often suggests that the expression might be the result of squaring a binomial involving a square root.

By rewriting the numerator as a perfect square, we were able to take the square root and simplify the expression to $\sin(\frac{x}{2}) = \frac{2 + \sqrt{5}}{3\sqrt{2}}$. However, we're not quite done yet! It's generally considered good practice to rationalize the denominator, which means getting rid of any square roots in the denominator.

To rationalize the denominator, we multiply both the numerator and the denominator by $\sqrt{2}$. This gives us:

$$\sin(\frac{x}{2}) = \frac{(2 + \sqrt{5})\sqrt{2}}{3\sqrt{2}\sqrt{2}} = \frac{2\sqrt{2} + \sqrt{10}}{6}$$

This is our final simplified answer. Notice that we no longer have any square roots in the denominator, and the expression is in its simplest form. Simplifying the solution not only makes the answer look neater but also makes it easier to work with in further calculations.

Common Mistakes to Avoid

Alright, let's talk about some common mistakes to avoid when tackling problems like this. Knowing what pitfalls to watch out for can save you a lot of headaches and help you nail the correct answer every time.

One of the most frequent errors is forgetting about the quadrants. Remember, the signs of trigonometric functions vary depending on the quadrant. For instance, in this problem, we knew that $x$ was in the second quadrant, which meant that $\cos(x)$ was negative. If we had forgotten this and taken the positive root for $\cos(x)$, we would have ended up with the wrong answer. So, always pay close attention to the given range of the angle and consider the signs of the trigonometric functions in that quadrant. This is super, super important, guys!

Another common mistake is messing up the half-angle formula. It's crucial to remember the correct formula and the "$\pm$" sign in front of the square root. You need to determine the correct sign based on the quadrant of $x/2$. In our case, $x/2$ was in the first quadrant, so we took the positive root. But if $x/2$ were in a different quadrant, we might need to take the negative root. Double-checking the quadrant and the sign is always a good idea.

Algebraic errors are another potential pitfall. When simplifying expressions with square roots and fractions, it's easy to make mistakes. Take your time, write down each step clearly, and double-check your work. Especially when you're dealing with nested fractions or expressions like $\sqrt{\frac{9 + 4\sqrt{5}}{18}}$, it's crucial to be careful and methodical.

Finally, don't forget to simplify your answer as much as possible. As we discussed earlier, simplifying the solution not only makes it look neater but also makes it easier to work with. Make sure to rationalize the denominator if necessary and look for opportunities to express the answer in its simplest form. This often involves recognizing perfect squares or other algebraic patterns.

By being aware of these common mistakes and taking steps to avoid them, you'll be well on your way to mastering trigonometric problems like this one. Remember, practice makes perfect, so keep working through problems and refining your skills!

Conclusion

So, there you have it, guys! We've successfully navigated this trigonometric problem, from understanding the initial conditions to arriving at the final simplified solution. The journey through this problem highlights the importance of a few key concepts in trigonometry: understanding the relationships between trigonometric functions, utilizing trigonometric identities, and paying attention to the signs of functions in different quadrants.

We started by recognizing that $\csc(x) = 9$ implies $\sin(x) = 1/9$ and that the given range $90^{\circ} < x < 180^{\circ}$ places $x$ in the second quadrant. This crucial piece of information allowed us to determine the sign of $\cos(x)$ later on.

Then, we skillfully employed the Pythagorean identity to find $\cos(x)$, remembering to choose the negative root because cosine is negative in the second quadrant. With $\cos(x)$ in hand, we confidently applied the half-angle formula for sine, again carefully considering the sign based on the quadrant of $x/2$.

Finally, we demonstrated the importance of simplifying the solution. We recognized that the expression under the square root could be rewritten as a perfect square, and we rationalized the denominator to arrive at our final answer: $\sin(\frac{x}{2}) = \frac{2\sqrt{2} + \sqrt{10}}{6}$.

This problem serves as a fantastic example of how different trigonometric concepts come together to solve a seemingly complex problem. By breaking it down into manageable steps and paying attention to detail, we were able to arrive at the solution with confidence.

Remember, guys, practice is key! The more you work through problems like this, the more comfortable you'll become with trigonometric identities and techniques. So, keep practicing, keep exploring, and keep having fun with math! You've got this!