How To Find The Exact Value Of Tan(23π/12) Using Angle Formulas

Hey guys! Today, we're diving into a fun trigonometric problem: finding the exact value of $\tan \left(\frac{23 \pi}{12}\right)$. This might seem intimidating at first, but don't worry! We're going to break it down using angle addition or subtraction formulas. These formulas are super handy for dealing with angles that aren't on our standard unit circle, and $\frac{23 \pi}{12}$ definitely falls into that category. So, grab your favorite beverage, and let's get started!

Understanding the Angle

First things first, let's get a good grasp of the angle $\frac{23 \pi}{12}$. It's not one we commonly encounter directly on the unit circle, like $\frac{\pi}{6}$, $\frac{\pi}{4}$, or $\frac{\pi}{3}$. To tackle this, we need to think about how we can express it as a sum or difference of angles that we do know. Remember, the unit circle is our friend here, and it holds the key to many trigonometric values. We need to rewrite $\frac{23 \pi}{12}$ into more manageable pieces. The goal is to find two angles that, when added or subtracted, give us $\frac{23 \pi}{12}$, and whose tangent values we already know or can easily calculate. For example, we might think about breaking it down into fractions with denominators that are common on the unit circle, like 3, 4, or 6. This way, we can leverage our knowledge of special angles like 30, 45, and 60 degrees (or their radian equivalents).

When dealing with angles like $\frac{23 \pi}{12}$, it's often helpful to think about the equivalent angle within a single revolution around the unit circle (i.e., between 0 and $2\pi$). Since $\frac{23 \pi}{12}$ is less than $2\pi$ (which is $\frac{24 \pi}{12}$), we don't need to subtract any multiples of $2\pi$. This makes our task a bit simpler. Now, let's explore some ways to break down this fraction into sums or differences of simpler fractions. We could try expressing it as a sum of two fractions, each with a denominator of 12, or perhaps look for combinations with denominators like 3, 4, and 6 that might simplify things. The key is to experiment and see what works best. By understanding the angle and its relationship to the unit circle, we set ourselves up for success in applying the angle addition or subtraction formulas.

Decomposing the Angle into Familiar Values

The main objective now is to express $\frac{23 \pi}{12}$ as a sum or difference of angles whose tangent values we know. Let's explore some possibilities. We can rewrite $\frac{23 \pi}{12}$ as $\frac{20 \pi}{12} + \frac{3 \pi}{12}$. Simplifying these fractions, we get $\frac{5 \pi}{3} + \frac{\pi}{4}$. Aha! Both $\frac{5 \pi}{3}$ and $\frac{\pi}{4}$ are angles we recognize from the unit circle. Alternatively, we could express $\frac{23 \pi}{12}$ as a difference. How about $\frac{2 \pi}{12}$ less than $2\pi$? That gives us $2\pi - \frac{\pi}{12}$, but $\frac{\pi}{12}$ isn't a standard angle. Let’s try another approach. Notice that $\frac{23 \pi}{12}$ is close to $2\pi$, so let's think about subtracting a small angle from $2\pi$. If we subtract $\frac{\pi}{12}$ from $2\pi$ (which is $\frac{24 \pi}{12}$), we get our target angle. However, $\frac{\pi}{12}$ isn't a standard angle, so this approach might not be the most straightforward. Another way to decompose the angle is to look for fractions that add up to $\frac{23}{12}$. We could try splitting it into two fractions, say $\frac{A}{B} + \frac{C}{D} = \frac{23}{12}$, where $B$ and $D$ are denominators that are common in the unit circle (like 3, 4, or 6). For example, we could consider combinations like $\frac{\pi}{3}$ and $\frac{\pi}{4}$, or $\frac{\pi}{6}$ and $\frac{\pi}{4}$. Let's see if any of these combinations work. If we use $\frac{5 \pi}{3}$ and $\frac{\pi}{4}$, we can proceed with the tangent addition formula. This decomposition seems promising because we know the tangent values of both $\frac{5 \pi}{3}$ and $\frac{\pi}{4}$.

Applying the Tangent Addition Formula

Now comes the fun part: using the tangent addition formula. The tangent addition formula states:

$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

In our case, we have $\tan \left(\frac{23 \pi}{12}\right) = \tan \left(\frac{5 \pi}{3} + \frac{\pi}{4}\right)$. So, let's identify our A and B: A = $\frac{5 \pi}{3}$ and B = $\frac{\pi}{4}$.

Before we plug these values into the formula, we need to find the tangent of each angle individually. Let's start with $\tan \left(\frac{5 \pi}{3}\right)$. Recall that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. For $\frac{5 \pi}{3}$, the reference angle is $\frac{\pi}{3}$, which has a tangent of $\sqrt{3}$. Since $\frac{5 \pi}{3}$ is in the fourth quadrant, where tangent is negative, we have $\tan \left(\frac{5 \pi}{3}\right) = -\sqrt{3}$. Next, let's find $\tan \left(\frac{\pi}{4}\right)$. This is a standard angle, and we know that $\tan \left(\frac{\pi}{4}\right) = 1$. Now that we have the tangent values for A and B, we can plug them into the tangent addition formula. So, we'll substitute $-\sqrt{3}$ for $\tan A$ and 1 for $\tan B$ in the formula. This will give us an expression that we can simplify to find the exact value of $\tan \left(\frac{23 \pi}{12}\right)$. Let's carefully substitute these values and then simplify step by step to avoid any errors. This is where the magic happens, and we'll see how the formula helps us unravel this trigonometric puzzle.

Substituting the values into the formula, we get:

$\tan \left(\frac{23 \pi}{12}\right) = \frac{-\sqrt{3} + 1}{1 - (-\sqrt{3})(1)} = \frac{1 - \sqrt{3}}{1 + \sqrt{3}}$

Simplifying the Expression

We've arrived at the expression $\frac{1 - \sqrt{3}}{1 + \sqrt{3}}$ for $\tan \left(\frac{23 \pi}{12}\right)$. While this is technically correct, it's common practice to rationalize the denominator, meaning we want to eliminate the square root from the denominator. To do this, we'll multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of $1 + \sqrt{3}$ is $1 - \sqrt{3}$. This clever trick allows us to get rid of the square root in the denominator because when we multiply a binomial by its conjugate, we get a difference of squares, which eliminates the radical.

So, let's multiply both the numerator and the denominator by $1 - \sqrt{3}$:

$\frac{1 - \sqrt{3}}{1 + \sqrt{3}} \cdot \frac{1 - \sqrt{3}}{1 - \sqrt{3}} = \frac{(1 - \sqrt{3})^2}{(1 + \sqrt{3})(1 - \sqrt{3})}$

Now, we need to expand both the numerator and the denominator. Let's start with the numerator. We have $(1 - \sqrt{3})^2$, which is the same as $(1 - \sqrt{3})(1 - \sqrt{3})$. Using the FOIL method (First, Outer, Inner, Last) or the binomial square formula, we get:

$(1 - \sqrt{3})(1 - \sqrt{3}) = 1 - \sqrt{3} - \sqrt{3} + 3 = 4 - 2\sqrt{3}$

Next, let's expand the denominator. We have $(1 + \sqrt{3})(1 - \sqrt{3})$, which is a difference of squares. This simplifies to:

$(1 + \sqrt{3})(1 - \sqrt{3}) = 1^2 - (\sqrt{3})^2 = 1 - 3 = -2$

Now, let's put the simplified numerator and denominator back into our expression:

$\frac{4 - 2\sqrt{3}}{-2}$

We can simplify this fraction further by dividing both terms in the numerator by -2:

$\frac{4 - 2\sqrt{3}}{-2} = -2 + \sqrt{3}$

Therefore, $\tan \left(\frac{23 \pi}{12}\right) = -2 + \sqrt{3}$.

Final Answer

Alright guys, we made it! We successfully found the exact value of $\tan \left(\frac{23 \pi}{12}\right)$ using the tangent addition formula and some algebraic manipulation. Our final answer is:

$\tan \left(\frac{23 \pi}{12}\right) = -2 + \sqrt{3}$

This problem highlights the power of using trigonometric identities and formulas to solve seemingly complex problems. By breaking down the angle into familiar components and applying the appropriate formula, we were able to arrive at a precise answer. Remember, practice makes perfect, so keep exploring these types of problems to strengthen your trigonometry skills. And remember, the unit circle is your best friend!