Simplify $\sqrt{\frac{1-cos 155}{1+cos 155}}$: A Trig Guide

Hey everyone! Today, we're diving into a fun little trigonometric problem that might look a bit intimidating at first, but trust me, we'll break it down together and simplify it like pros. We're going to tackle the expression $\sqrt{\frac{1-\cos 155^{\circ}}{1+\cos 155^{\circ}}}$. So, grab your thinking caps, and let's get started!

Understanding the Problem

Before we jump into the solution, let's make sure we understand what we're dealing with. We have a square root encompassing a fraction, and inside that fraction, we have cosine of 155 degrees. Now, 155 degrees isn't one of those angles we typically memorize (like 30, 45, or 60 degrees), so we'll need to use some trigonometric identities to help us out. The key here is to simplify this complex expression into something more manageable and, ideally, something we can directly calculate.

Trigonometric identities are your best friends in scenarios like these. They are like the Swiss Army knives of trigonometry, offering various ways to rewrite and simplify expressions. For this particular problem, we'll be leaning heavily on the half-angle identities. These identities are super useful when you have an angle that isn't a standard one, but you can relate it to a standard angle (in our case, we'll see how 155 degrees relates to something more familiar).

Now, let's talk about the cosine function itself. Cosine is one of the fundamental trigonometric functions, and it gives us the x-coordinate of a point on the unit circle corresponding to a given angle. Understanding how cosine behaves in different quadrants is crucial. 155 degrees lies in the second quadrant, where cosine values are negative. This is an important detail that will help us avoid sign errors later on. Remember, a solid understanding of the unit circle and the behavior of trigonometric functions in different quadrants is the backbone of solving these problems.

Finally, let's consider the square root. The square root function gives us the non-negative value that, when squared, equals the expression inside the root. This means we need to be mindful of the signs when we simplify our expression. We want to ensure that the final result is positive, as the square root of a number is always non-negative. So, we'll need to keep track of the signs of our trigonometric functions and make sure everything comes together correctly in the end.

Applying Half-Angle Identities

Okay, guys, this is where the magic happens! We're going to use the half-angle identities to simplify our expression. Remember those Swiss Army knives we talked about? Well, the half-angle identities are some of the sharpest blades in the set.

The half-angle identities for tangent are particularly useful here. There are a couple of ways to express the half-angle identity for tangent, but the one that fits our expression perfectly is: $\tan(\frac{x}{2}) = \sqrt{\frac{1-\cos x}{1+\cos x}}$. See the resemblance to our original problem? This is not a coincidence! This identity is tailor-made for simplifying expressions like ours. Now, you might be wondering, where did this identity come from? Well, it's derived from other fundamental trigonometric identities, but for now, let's focus on how to use it effectively.

So, how does this identity help us? Notice that our expression $\sqrt{\frac{1-\cos 155^{\circ}}{1+\cos 155^{\circ}}}$ looks exactly like the right-hand side of the half-angle identity! This means we can rewrite our expression in terms of a tangent function. We can set $x = 155^{\circ}$, and then our expression becomes $\tan(\frac{155^{\circ}}{2})$. This is a huge step forward because we've transformed a complex expression with a square root and a fraction into a single tangent function.

But we're not done yet! We still need to figure out what $\tan(\frac{155^{\circ}}{2})$ actually is. To do this, we need to calculate $\frac{155^{\circ}}{2}$, which is 77.5 degrees. Now, 77.5 degrees isn't a standard angle either, but it's a bit more manageable than 155 degrees. We've successfully reduced the complexity of the angle we're dealing with. This is a common strategy in trigonometry: use identities to break down complex angles into simpler ones.

So, now we have $\tan(77.5^{\circ})$. The next step is to see if we can simplify this further, perhaps by relating it to angles we know or by using another identity. Remember, the goal is to get to a value we can actually calculate or express in a simpler form. The half-angle identity was our first major tool, and now we'll need to see what other tricks we have up our sleeves to continue simplifying.

Calculating the Tangent Value

Alright, let's figure out the value of $\tan(77.5^{\circ})$. We know that 77.5 degrees is half of 155 degrees, and we used the half-angle identity to get to this point. However, 77.5 degrees itself isn't a standard angle, so we can't just look up its tangent value on a unit circle or in a table. We need to get a bit creative here.

One approach we can take is to think about angles that are close to 77.5 degrees and see if we can relate them in any way. For example, we know that $77.5^{\circ} = 45^{\circ} + 32.5^{\circ}$. While 32.5 degrees isn't a standard angle either, breaking it down this way might spark some ideas. We could potentially use the tangent addition formula, which states that $\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$. If we could find a way to express $\tan(32.5^{\circ})$ in a simpler form, we might be able to use this formula to calculate $\tan(77.5^{\circ})$.

Another approach is to go back to the half-angle identity. We used it to go from 155 degrees to 77.5 degrees, but maybe we can use it again in reverse! Think about it: if 77.5 degrees is half of 155 degrees, then maybe there's an angle we can find whose half-angle is 77.5 degrees. This might sound a bit confusing, but the idea is to find an angle $y$ such that $\frac{y}{2} = 77.5^{\circ}$. Solving for $y$, we get $y = 155^{\circ}$. Wait a minute... that's where we started! This might seem like we're going in circles, but it actually gives us a new perspective.

Instead of thinking about $\tan(77.5^{\circ})$ directly, let's think about its relationship to $\cos(155^{\circ})$. We know that $\tan(77.5^{\circ}) = \sqrt{\frac{1-\cos 155^{\circ}}{1+\cos 155^{\circ}}}$. The key here is to find the value of $\cos(155^{\circ})$. Now, 155 degrees is in the second quadrant, so its cosine will be negative. We can relate 155 degrees to its reference angle, which is $180^{\circ} - 155^{\circ} = 25^{\circ}$. So, $\cos(155^{\circ}) = -\cos(25^{\circ})$.

Unfortunately, 25 degrees isn't a standard angle either, and finding the exact value of $\cos(25^{\circ})$ requires more advanced techniques or a calculator. However, we've made significant progress in simplifying the problem. We've reduced it to finding the cosine of 25 degrees, which is a specific value that can be looked up or approximated. The key takeaway here is that even if we can't find an exact, simple answer, we've still simplified the expression significantly by using trigonometric identities and reasoning about angles and quadrants.

The Final Simplified Form

Okay, let's recap where we are. We started with $\sqrt{\frac{1-\cos 155^{\circ}}{1+\cos 155^{\circ}}}$ and used the half-angle identity for tangent to simplify it to $\tan(77.5^{\circ})$. We then realized that 77.5 degrees isn't a standard angle, but we could relate it back to $\cos(155^{\circ})$. We found that $\cos(155^{\circ}) = -\cos(25^{\circ})$, and we know that $\tan(77.5^{\circ}) = \sqrt{\frac{1-\cos 155^{\circ}}{1+\cos 155^{\circ}}}$.

Now, let's plug in $\cos(155^{\circ}) = -\cos(25^{\circ})$ into our expression for $\tan(77.5^{\circ})$: $\tan(77.5^{\circ}) = \sqrt{\frac{1-(-\cos 25^{\circ})}{1+(-\cos 25^{\circ})}} = \sqrt{\frac{1+\cos 25^{\circ}}{1-\cos 25^{\circ}}}$. This is a simplified form, but we can actually go a bit further.

Remember that we're trying to find a simple, clean expression. Having square roots and fractions inside the square root isn't ideal. Let's try to rationalize the denominator inside the square root. To do this, we multiply both the numerator and the denominator by the conjugate of the denominator, which is $(1 + \cos 25^{\circ})$:

$\tan(77.5^{\circ}) = \sqrt{\frac{(1+\cos 25^{\circ})(1+\cos 25^{\circ})}{(1-\cos 25^{\circ})(1+\cos 25^{\circ})}} = \sqrt{\frac{(1+\cos 25^{\circ})^2}{1-\cos^2 25^{\circ}}}$.

Now, we can use another fundamental trigonometric identity: $\sin^2 x + \cos^2 x = 1$. This means that $1 - \cos^2 x = \sin^2 x$. So, we can replace the denominator with $\sin^2 25^{\circ}$:

$\tan(77.5^{\circ}) = \sqrt{\frac{(1+\cos 25^{\circ})^2}{\sin^2 25^{\circ}}}$.

Now we have a perfect square in both the numerator and the denominator! We can take the square root of both:

$\tan(77.5^{\circ}) = \frac{1+\cos 25^{\circ}}{\sin 25^{\circ}}$.

This is a much cleaner expression! We've gotten rid of the nested square root and simplified the fraction. We can even break this down further into two separate fractions: $\tan(77.5^{\circ}) = \frac{1}{\sin 25^{\circ}} + \frac{\cos 25^{\circ}}{\sin 25^{\circ}}$.

We know that $\frac{1}{\sin x} = \csc x$ (cosecant) and $\frac{\cos x}{\sin x} = \cot x$ (cotangent). So, we can rewrite our expression as:

$\tan(77.5^{\circ}) = \csc 25^{\circ} + \cot 25^{\circ}$.

And there you have it! We've simplified the original expression $\sqrt{\frac{1-\cos 155^{\circ}}{1+\cos 155^{\circ}}}$ to $\csc 25^{\circ} + \cot 25^{\circ}$. While we still have trigonometric functions of 25 degrees, this is a much simpler form than where we started. It highlights the power of trigonometric identities and how they can be used to transform complex expressions into more manageable ones.

Conclusion

So, guys, we made it! We successfully simplified a seemingly complex trigonometric expression by using half-angle identities, rationalizing denominators, and applying fundamental trigonometric relationships. We saw how crucial it is to understand trigonometric identities and how they act as powerful tools in simplifying expressions. We also emphasized the importance of knowing the behavior of trigonometric functions in different quadrants and how this knowledge can help avoid errors.

Remember, the key to mastering trigonometry is practice. The more problems you solve, the more comfortable you'll become with recognizing patterns and applying the right identities. Don't be afraid to experiment and try different approaches. And most importantly, have fun with it! Trigonometry can be a fascinating subject, and the feeling of successfully simplifying a complex expression is definitely worth the effort. Keep practicing, and you'll be a trig whiz in no time!