Simplify Sin 6° Cos 12° Cos 24° Cos 48°

Hey math whizzes and curious minds! Today, we're diving deep into the world of trigonometry to tackle a problem that might look a little intimidating at first glance: simplifying the expression sin 6° cos 12° cos 24° cos 48°. We've all been there, staring at a long string of trigonometric functions and wondering, "Where do I even begin?" Well, buckle up, because we're going to break this down step-by-step, making it super clear and easy to follow. You'll be impressing your friends with your trigonometric prowess in no time!

The Power of Double Angle Formulas

When you see a series of cosines with angles doubling each time, like we have with 12°, 24°, and 48°, a little bell should go off in your head. This usually signals that the double angle formula for sine is our best friend. Remember this gem? It states that sin(2θ) = 2 sin(θ) cos(θ). This formula is the key to unlocking this problem. We can rearrange it to cos(θ) = sin(2θ) / (2 sin(θ)). This little trick allows us to convert a cosine term into a sine term, which can simplify things dramatically. We'll be using this concept repeatedly to chip away at our expression. The goal is to strategically introduce sine terms that will cancel out with subsequent terms, leaving us with a much simpler result. It's like a trigonometric puzzle, and we've just found the first piece!

Let's Get Our Hands Dirty!

Alright guys, let's start manipulating our expression: sin 6° cos 12° cos 24° cos 48°. The trick here is to multiply and divide by something that will help us use the double angle formula. Notice that we have a sin 6° term already. If we could somehow get a 2 sin 6° cos 6° in there, we could use the double angle formula to turn it into sin 12°. But we don't have cos 6°. So, what if we multiply the entire expression by cos 6° and then immediately divide by it? That keeps the value the same, right? But that seems to complicate things. Instead, let's try multiplying the entire expression by 2 sin 6° and then dividing by 2 sin 6° outside the expression. This might seem a bit backward, but trust me, it works wonders.

So, we have:

(1 / (2 sin 6°)) * (2 sin 6° cos 12° cos 24° cos 48°)

This doesn't seem to help directly because the first sine term is sin 6°, not cos 6°. Let's try a different approach. What if we multiply and divide by cos 6°? That might seem to add complexity.

Let's reconsider the double angle formula: sin(2θ) = 2 sin(θ) cos(θ). We want to create pairs of sin(θ) cos(θ). Our current expression starts with sin 6°. If we had cos 6°, we could make sin 12°. But we don't. Instead, we have cos 12°. This suggests we should look for a sin 12° to pair with.

Let's take our original expression: $E = \sin 6° \cos 12° \cos 24° \cos 48°$.

What if we multiply the entire expression by $2 \sin 6°$ and then divide by $2 \sin 6°$? This gives us:

$E = \frac{1}{2 \sin 6°} (2 \sin 6° \cos 12° \cos 24° \cos 48°)$

This doesn't seem right because we don't have a cos 6° to pair with sin 6°. Let's try a different angle. The angles are 6°, 12°, 24°, 48°. They are doubling. The initial angle is 6°.

Let's rewrite the expression as $E = \sin 6° \cos 12° \cos 24° \cos 48°$.

We want to use the identity $2 \sin A \cos A = \sin 2A$. Notice that the angles are 12°, 24°, 48°, which are $2 imes 6°$, $2 imes 12°$, $2 imes 24°$. This structure is a huge hint.

Let's try multiplying the entire expression by $2 \cos 6°$. Then we'd divide by $2 \cos 6°$ to keep it the same.

$E = \frac{1}{2 \cos 6°} (2 \cos 6° \sin 6° \cos 12° \cos 24° \cos 48°)$

Now, using $2 \sin A \cos A = \sin 2A$ with $A = 6°$, we get $2 \sin 6° \cos 6° = \sin(2 \times 6°) = \sin 12°$. So our expression becomes:

$E = \frac{1}{2 \cos 6°} (\sin 12° \cos 12° \cos 24° \cos 48°)$

Now, look at the part in the parentheses: $\sin 12° \cos 12°$. If we multiply this by 2, we get $\sin 24°$. So, let's multiply the entire expression by 2 and divide by 2 again. This means multiplying by $1$ effectively. It's easier to see if we carry the denominator along.

Let's restart with the original expression $E = \sin 6° \cos 12° \cos 24° \cos 48°$.

We want to use the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$.

Multiply and divide by $2 \cos 6°$:

$E = \frac{2 \cos 6° \sin 6° \cos 12° \cos 24° \cos 48°}{2 \cos 6°}$

Using the double angle formula $2 \sin 6° \cos 6° = \sin(2 \times 6°) = \sin 12°$:

$E = \frac{\sin 12° \cos 12° \cos 24° \cos 48°}{2 \cos 6°}$

Now, we have $\sin 12° \cos 12°$. Multiply the numerator and denominator by 2:

$E = \frac{2 \sin 12° \cos 12° \cos 24° \cos 48°}{2 \times (2 \cos 6°)}$

Using the double angle formula again, $2 \sin 12° \cos 12° = \sin(2 \times 12°) = \sin 24°$:

$E = \frac{\sin 24° \cos 24° \cos 48°}{4 \cos 6°}$

We're on a roll! Next, we have $\sin 24° \cos 24°$. Multiply the numerator and denominator by 2:

$E = \frac{2 \sin 24° \cos 24° \cos 48°}{2 \times (4 \cos 6°)}$

Using the double angle formula, $2 \sin 24° \cos 24° = \sin(2 \times 24°) = \sin 48°$:

$E = \frac{\sin 48° \cos 48°}{8 \cos 6°}$

And finally, we have $\sin 48° \cos 48°$. Multiply the numerator and denominator by 2:

$E = \frac{2 \sin 48° \cos 48°}{2 \times (8 \cos 6°)}$

Using the double angle formula one last time, $2 \sin 48° \cos 48° = \sin(2 \times 48°) = \sin 96°$:

$E = \frac{\sin 96°}{16 \cos 6°}$

The Final Stretch: Unveiling the Simplification

We're almost there, guys! We have the expression $\frac{\sin 96°}{16 \cos 6°}$. Now, we need to see if $\sin 96°$ and $\cos 6°$ are related. Remember that sine and cosine are complementary functions, meaning $\sin \theta = \cos(90° - \theta)$ and $\cos \theta = \sin(90° - \theta)$. This is a crucial identity that often pops up in these types of problems. We can use this to relate $\sin 96°$ to a cosine function.

Let's look at $\sin 96°$. We know that $\sin(180° - \theta) = \sin \theta$. So, $\sin 96° = \sin(180° - 96°) = \sin 84°$. This doesn't seem to help directly with $\cos 6°$.

Let's use the complementary angle identity: $\sin \theta = \cos(90° - \theta)$.

So, $\sin 96° = \cos(90° - 96°) = \cos(-6°)$.

Since the cosine function is an even function, meaning $\cos(-\theta) = \cos \theta$, we have $\cos(-6°) = \cos 6°$.

Therefore, $\sin 96° = \cos 6°$.

Now we can substitute this back into our expression for E:

$E = \frac{\cos 6°}{16 \cos 6°}$

And just like that, the $\cos 6°$ terms cancel out!

$E = \frac{1}{16}$

Conclusion: A Simple Answer from a Complex Expression

So there you have it! The expression $\sin 6° \cos 12° \cos 24° \cos 48°$ simplifies to the surprisingly simple fraction 1/16. Pretty neat, right? We started with a bunch of trigonometric terms and ended up with a clean, numerical answer by strategically applying the double angle formula for sine and the complementary angle identity. These are fundamental tools in trigonometry, and seeing them work together like this is really satisfying. Remember, when you see angles doubling in a product like this, the double angle formula is often your key. Don't be afraid to multiply and divide by clever terms to make those identities work for you. Keep practicing these techniques, and you'll find that even the most complex-looking trigonometric problems can be solved with a bit of strategy and the right identities. Keep exploring the amazing world of math, guys!