Find Cos(59π/12): Exact Value With Angle Formulas

Dec 15, 2025 by Andrew McMorgan 50 views

Hey math enthusiasts! Today, we're diving deep into the fascinating world of trigonometry to tackle a rather imposing-looking problem: finding the exact value of cos(59π/12). Now, I know what you might be thinking – "59π/12? That sounds like a mouthful!" But don't worry, guys, with the right tools, specifically the angle sum or difference formulas, this seemingly complex expression will unravel into a straightforward calculation. We're going to break down this problem step-by-step, making sure you understand the 'why' behind each move. Get ready to flex those mathematical muscles because this is where the magic of trigonometry truly shines, allowing us to pinpoint precise values for angles that aren't our usual suspects like 30°, 45°, or 60°.

Deconstructing the Angle: The First Crucial Step

Before we can even think about applying any formulas, the very first thing we need to do is deconstruct the angle 59π/12. The key here is to express this angle as a sum or difference of two familiar angles for which we already know the exact trigonometric values. Our usual suspects are angles like π/6, π/4, π/3, π/2, and their multiples. The goal is to manipulate 59π/12 so it fits the form $A pm B$ , where $A$ and $B$ are angles whose sine and cosine values we can readily recall. Let's try to find a common denominator for our usual angles. A good common denominator for our standard angles (like 3, 4, 6, 2) is 12. So, let's express our standard angles with a denominator of 12:

π/6 = 2π/12
π/4 = 3π/12
π/3 = 4π/12
π/2 = 6π/12

Now, we need to see if we can combine these to get 59π/12. It might seem a bit daunting at first, but often, it's helpful to think about angles slightly larger than 2π (which is 24π/12). Let's consider some multiples of π/3 and π/4.

Option 1: Using π/3 and π/4. We know that 59 is a large number. Let's try multiples of 4π/12 (π/3) and 3π/12 (π/4). What if we take a large multiple of, say, π/3? $10 imes rac{\pi}{3} = rac{40 iu}{12}$ . This is not quite 59π/12. Let's try multiples of π/4: $15 imes rac{\pi}{4} = rac{45 iu}{12}$ . Still not there. Consider $2 iu = rac{24 iu}{12}$ . Maybe we can express $59 iu/12$ as something related to $2 iu$ ? $59 iu/12$ is greater than $2 iu$ . In fact, $59 iu/12 = 4 iu + 7 iu/12$ . This doesn't seem helpful immediately.

Let's try a different approach. We can also think about angles that are close to 59π/12. Since 59 is close to 60, let's consider 60π/12, which simplifies to 5π. That's 5 full rotations, which brings us back to 0. This isn't quite right. How about $59 iu/12$ ? It's close to $60 iu/12 = 5 iu$ . It's also close to $48 iu/12 = 4 iu$ .

Let's look at multiples of π/12. We know that $rac{ iu}{12}$ is $15^ iu$ . So $59 iu/12$ is $(59 imes 15)^ iu = 885^ iu$ . This is a very large angle! We can subtract multiples of $360^ iu$ (or $2 iu$ ) to find a coterminal angle in the range $[0, 2 iu)$ .

$885^ iu - 2 imes 360^ iu = 885^ iu - 720^ iu = 165^ iu$ . So, $rac{59 iu}{12}$ is coterminal with $165^ iu$ . Now, $165^ iu$ is an angle we can work with! We can express $165^ iu$ as a sum or difference of familiar angles. For instance:

$165^ iu = 120^ iu + 45^ iu$ (which is $rac{2 iu}{3} + rac{ iu}{4}$ )
$165^ iu = 135^ iu + 30^ iu$ (which is $rac{3 iu}{4} + rac{ iu}{6}$ )
$165^ iu = 180^ iu - 15^ iu$ (which is $iu - rac{ iu}{12}$ )

Let's stick with radians for now. We found that $59 iu/12$ is coterminal with $165^ iu$ . In radians, $165^ iu = 165 imes rac{ iu}{180} = rac{11 iu}{12}$ .

So, we need to find $rac{11 iu}{12}$ . Let's express $rac{11 iu}{12}$ as a sum or difference of angles whose trig values we know. Possible combinations include:

$rac{11 iu}{12} = rac{8 iu}{12} + rac{3 iu}{12} = rac{2 iu}{3} + rac{ iu}{4}$
$rac{11 iu}{12} = rac{9 iu}{12} + rac{2 iu}{12} = rac{3 iu}{4} + rac{ iu}{6}$
$rac{11 iu}{12} = rac{12 iu}{12} - rac{ iu}{12} = iu - rac{ iu}{12}$ . This one looks promising if we can handle $rac{ iu}{12}$ , but finding its exact value might require another step. Let's try the first two.

Let's use $rac{11 iu}{12} = rac{2 iu}{3} + rac{ iu}{4}$ . We know the exact values for $rac{2 iu}{3}$ and $rac{ iu}{4}$ .

$rac{2 iu}{3}$ is $120^ iu$ . $rac{ iu}{4}$ is $45^ iu$ . $120^ iu + 45^ iu = 165^ iu$ . This works perfectly.

So, we have successfully broken down $rac{59 iu}{12}$ into a sum of two angles whose trigonometric values are known: $rac{2 iu}{3}$ and $rac{ iu}{4}$ . This is our crucial first step!

Applying the Angle Sum Formula for Cosine

Now that we've established $rac{59 iu}{12}$ is coterminal with $rac{11 iu}{12}$ , and that $rac{11 iu}{12} = rac{2 iu}{3} + rac{ iu}{4}$ , we can deploy the angle sum formula for cosine. This formula states:

$iu pm (A + B) = iu A iu B - iu A iu B$

In our case, $A = rac{2 iu}{3}$ and $B = rac{ iu}{4}$ . So, we need to calculate:

$iu pm group( rac{2 iu}{3} + rac{ iu}{4} group) = iu group( rac{2 iu}{3} group) iu group( rac{ iu}{4} group) - iu group( rac{2 iu}{3} group) iu group( rac{ iu}{4} group)$

To do this, we need the exact values of sine and cosine for both $rac{2 iu}{3}$ and $rac{ iu}{4}$ . Let's recall them:

For $rac{2 iu}{3}$ (which is $120^ iu$ ):
- $iu group( rac{2 iu}{3} group) = - rac{1}{2}$
- $iu group( rac{2 iu}{3} group) = rac{\sqrt{3}}{2}$
For $rac{ iu}{4}$ (which is $45^ iu$ ):
- $iu group( rac{ iu}{4} group) = rac{\sqrt{2}}{2}$
- $iu group( rac{ iu}{4} group) = rac{\sqrt{2}}{2}$

Now, let's substitute these values into the angle sum formula:

$iu pm group( rac{2 iu}{3} + rac{ iu}{4} group) = group( - rac{1}{2} group) group( rac{\sqrt{2}}{2} group) - group( rac{\sqrt{3}}{2} group) group( rac{\sqrt{2}}{2} group)$

$iu pm group( rac{2 iu}{3} + rac{ iu}{4} group) = - rac{\sqrt{2}}{4} - rac{\sqrt{6}}{4}$

$iu pm group( rac{2 iu}{3} + rac{ iu}{4} group) = - rac{\sqrt{2} + \sqrt{6}}{4}$

So, the exact value of $iu pm group( rac{59 iu}{12} group)$ is $- rac{\sqrt{2} + \sqrt{6}}{4}$ . Pretty neat, right? This confirms our coterminal angle approach was sound and the formula application was correct. This is the power of breaking down complex problems into manageable parts using the fundamental identities we learn in trigonometry.

Alternative Approach: Using the Angle Difference Formula

Just to show you another way and reinforce the concept, let's see if we can use an angle difference formula. Remember, $rac{59 iu}{12}$ is coterminal with $rac{11 iu}{12}$ . We could also express $rac{11 iu}{12}$ using a difference. For example:

$rac{11 iu}{12} = rac{3 iu}{4} - rac{ iu}{6}$ is NOT correct, because $rac{9 iu}{12} - rac{2 iu}{12} = rac{7 iu}{12}$ .

Let's try $rac{11 iu}{12} = rac{4 iu}{3} - rac{5 iu}{12}$ ? No, let's find simpler angles.

How about $rac{11 iu}{12} = rac{5 iu}{6} - rac{ iu}{12}$ ? This is also not helpful as we don't know $rac{ iu}{12}$ easily.

Let's revisit our coterminal angle $165^ iu$ . We can express $165^ iu$ as $210^ iu - 45^ iu$ or $180^ iu - 15^ iu$ . The latter involves $15^ iu$ , which itself needs to be broken down. Let's use $210^ iu - 45^ iu$ for practice. In radians, this is $rac{7 iu}{6} - rac{ iu}{4}$ .

So, we want to calculate $iu pm group( rac{7 iu}{6} - rac{ iu}{4} group)$ . The angle difference formula for cosine is:

$iu pm (A - B) = iu A iu B + iu A iu B$

Here, $A = rac{7 iu}{6}$ and $B = rac{ iu}{4}$ . We need the values for $rac{7 iu}{6}$ and $rac{ iu}{4}$ .

For $rac{7 iu}{6}$ (which is $210^ iu$ ):
- $iu group( rac{7 iu}{6} group) = - rac{\sqrt{3}}{2}$
- $iu group( rac{7 iu}{6} group) = - rac{1}{2}$
For $rac{ iu}{4}$ (which is $45^ iu$ ):
- $iu group( rac{ iu}{4} group) = rac{\sqrt{2}}{2}$
- $iu group( rac{ iu}{4} group) = rac{\sqrt{2}}{2}$

Plugging these into the difference formula:

$iu pm group( rac{7 iu}{6} - rac{ iu}{4} group) = group( - rac{\sqrt{3}}{2} group) group( rac{\sqrt{2}}{2} group) + group( - rac{1}{2} group) group( rac{\sqrt{2}}{2} group)$

$iu pm group( rac{7 iu}{6} - rac{ iu}{4} group) = - rac{\sqrt{6}}{4} - rac{\sqrt{2}}{4}$

$iu pm group( rac{7 iu}{6} - rac{ iu}{4} group) = - rac{\sqrt{6} + \sqrt{2}}{4}$

This result is identical to the one we obtained using the angle sum formula! This is fantastic because it shows that as long as we correctly identify our component angles and apply the appropriate formula, we'll arrive at the correct exact value. It's like having multiple paths leading to the same beautiful destination.

The Significance of Exact Values

So, why do we bother with these