Solving $\frac{\sin(x)}{\cos(8^\circ-x)}=\frac{\cos(68^\circ)\sin(24^\circ)}{\cos(80^\circ)\cos(64^\circ)}$ A Trigonometric Journey

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Hey guys! Today, we're diving deep into solving a fascinating trigonometric equation. This isn't your everyday sine and cosine problem; it's a journey that involves clever trigonometric manipulations, strategic identity applications, and a dash of problem-solving flair. We're going to break down the equation sin⁡(x)cos⁡(8∘−x)=cos⁡(68∘)sin⁡(24∘)cos⁡(80∘)cos⁡(64∘)\frac{\sin(x)}{\cos(8^\circ-x)}=\frac{\cos(68^\circ)\sin(24^\circ)}{\cos(80^\circ)\cos(64^\circ)} for 0∘<x<90∘0^\circ<x<90^\circ. So, buckle up and let's get started!

Understanding the Problem

Before we jump into solutions, let's make sure we understand what we're dealing with. Trigonometric equations like this one can seem daunting at first. You've got sines, cosines, angles flying around, and fractions to boot! But don't worry, the key is to approach it systematically. Our goal is to find the value(s) of x that satisfy the equation within the given interval (0° < x < 90°). We'll be using our trusty toolbox of trigonometric identities – those sum-to-product, product-to-sum, angle addition/subtraction, and double-angle formulas – to simplify the equation and hopefully isolate x. Think of it like a puzzle; each identity is a piece that, when placed correctly, helps reveal the solution. And remember, precision is paramount. In trigonometry, a small error can throw everything off, so we'll be meticulous in each step.

Strategic Trigonometric Transformations

Alright, let’s kick things off by tackling the right-hand side of the equation. It looks a bit clunky, doesn't it? We have a product of cosines and sines, which screams for some product-to-sum identities. These identities are super handy for transforming products of trigonometric functions into sums or differences, often making the expression much easier to handle. Specifically, we'll be focusing on the identities:

  • 2cos(A)cos(B) = cos(A + B) + cos(A - B)
  • 2sin(A)cos(B) = sin(A + B) + sin(A - B)

Why these? Well, notice we have cos(68°)sin(24°), cos(80°), and cos(64°). These angles might seem random, but they’re hinting at a hidden structure. We’ll start by applying the product-to-sum identity to the denominator, cos(80°)cos(64°), and then see if we can simplify further using other identities. Remember, the beauty of trigonometry lies in recognizing these patterns and applying the right tools at the right time. It's like being a mathematical detective, piecing together clues to solve the mystery!

To make things clearer, let's rewrite the right-hand side (RHS) as:

cos⁡(68∘)sin⁡(24∘)cos⁡(80∘)cos⁡(64∘)=cos⁡(68∘)sin⁡(24∘)12[cos⁡(80∘+64∘)+cos⁡(80∘−64∘)]\frac{\cos(68^\circ)\sin(24^\circ)}{\cos(80^\circ)\cos(64^\circ)} = \frac{\cos(68^\circ)\sin(24^\circ)}{\frac{1}{2}[\cos(80^\circ + 64^\circ) + \cos(80^\circ - 64^\circ)]}

Simplifying this gives us:

2cos⁡(68∘)sin⁡(24∘)cos⁡(144∘)+cos⁡(16∘)\frac{2\cos(68^\circ)\sin(24^\circ)}{\cos(144^\circ) + \cos(16^\circ)}

Now, let's work on the numerator, 2cos(68°)sin(24°). We can use the product-to-sum identity 2sin(A)cos(B) = sin(A + B) + sin(A - B), but we'll need to swap the order of sine and cosine to match the identity. So, we rewrite it as 2sin(24°)cos(68°). Applying the identity, we get:

2sin(24°)cos(68°) = sin(24° + 68°) + sin(24° - 68°) = sin(92°) + sin(-44°)

Since sin(-x) = -sin(x), we have:

sin(92°) - sin(44°)

Also, remember that sin(180° - x) = sin(x), so sin(92°) = sin(180° - 92°) = sin(88°). Thus, our numerator becomes:

sin(88°) - sin(44°)

More Transformations and Simplifications

Okay, we've made some good progress simplifying the numerator and denominator. But we're not done yet! The expression sin(88°) - sin(44°) in the numerator looks like it's begging for a sum-to-product identity. These identities are the reverse of the product-to-sum ones, turning differences (or sums) of trigonometric functions into products. The specific identity we'll use is:

sin(A) - sin(B) = 2cos((A + B) / 2)sin((A - B) / 2)

Applying this to sin(88°) - sin(44°), we get:

2cos((88° + 44°) / 2)sin((88° - 44°) / 2) = 2cos(66°)sin(22°)

Now, let’s circle back to the denominator, which was cos(144°) + cos(16°). This also calls for a sum-to-product identity, this time for cosines:

cos(A) + cos(B) = 2cos((A + B) / 2)cos((A - B) / 2)

Applying this, we have:

cos(144°) + cos(16°) = 2cos((144° + 16°) / 2)cos((144° - 16°) / 2) = 2cos(80°)cos(64°)

Woah, hold on a second! Notice anything familiar about this denominator? It's the same as our original denominator on the RHS! This is a fantastic sign, as it means we're on the right track. Now, let’s put everything back together. The right-hand side of our equation now looks like this:

2cos(66∘)sin(22∘)2cos(80∘)cos(64∘)\frac{2cos(66^\circ)sin(22^\circ)}{2cos(80^\circ)cos(64^\circ)}

Simplifying by canceling the 2's, we have:

cos(66∘)sin(22∘)cos(80∘)cos(64∘)\frac{cos(66^\circ)sin(22^\circ)}{cos(80^\circ)cos(64^\circ)}

Back to the Original Equation

Okay, we've transformed the right-hand side into something much more manageable. Let's bring it back into our original equation:

sin⁡(x)cos⁡(8∘−x)=cos⁡(66∘)sin⁡(22∘)cos⁡(80∘)cos⁡(64∘)\frac{\sin(x)}{\cos(8^\circ-x)} = \frac{\cos(66^\circ)\sin(22^\circ)}{\cos(80^\circ)\cos(64^\circ)}

Now, this might still look intimidating, but we're getting closer. The key here is to realize that we need to relate the left-hand side (LHS) to the simplified RHS. We have a sine and a cosine on both sides, so maybe we can manipulate the LHS using some angle addition or subtraction identities. Let's focus on the denominator of the LHS, cos(8° - x). Using the cosine subtraction identity, cos(A - B) = cos(A)cos(B) + sin(A)sin(B), we get:

cos(8° - x) = cos(8°)cos(x) + sin(8°)sin(x)

So, our LHS becomes:

sin⁡(x)cos(8∘)cos(x)+sin(8∘)sin(x)\frac{\sin(x)}{cos(8^\circ)cos(x) + sin(8^\circ)sin(x)}

At this point, it's tempting to try cross-multiplying and simplifying further, but let's take a step back and look at the bigger picture. We need to find a value for x that makes the LHS equal to the simplified RHS. Is there a way to make the LHS look more like the RHS without blindly applying more identities?

Spotting the Solution

This is where the real problem-solving magic happens. Instead of just pushing symbols around, let's use a bit of insight. We have the equation:

sin⁡(x)cos⁡(8∘−x)=cos⁡(66∘)sin⁡(22∘)cos⁡(80∘)cos⁡(64∘)\frac{\sin(x)}{\cos(8^\circ-x)} = \frac{\cos(66^\circ)\sin(22^\circ)}{\cos(80^\circ)\cos(64^\circ)}

Notice the structure. On the LHS, we have sin(x) in the numerator and cos(8° - x) in the denominator. On the RHS, we have a fraction involving sines and cosines of specific angles. What if we could make the angles on the LHS match the angles on the RHS in some way? This is a classic technique in problem-solving: look for patterns and try to match structures.

Let's consider the numerator first. We have sin(x) on the LHS and sin(22°) on the RHS. Could x possibly be 22°? If x = 22°, then the LHS numerator would match one of the terms on the RHS. This is a great starting point. Now, let's see what happens to the denominator if x = 22°:

cos(8° - x) = cos(8° - 22°) = cos(-14°)

Since cos(-x) = cos(x), we have cos(-14°) = cos(14°). So, if x = 22°, the LHS denominator becomes cos(14°). Now our equation looks like this:

sin⁡(22∘)cos⁡(14∘)=cos⁡(66∘)sin⁡(22∘)cos⁡(80∘)cos⁡(64∘)\frac{\sin(22^\circ)}{\cos(14^\circ)} = \frac{\cos(66^\circ)\sin(22^\circ)}{\cos(80^\circ)\cos(64^\circ)}

We can cancel out sin(22°) from both sides (since it's not zero), leaving us with:

1cos⁡(14∘)=cos⁡(66∘)cos⁡(80∘)cos⁡(64∘)\frac{1}{\cos(14^\circ)} = \frac{\cos(66^\circ)}{\cos(80^\circ)\cos(64^\circ)}

Cross-multiplying, we get:

cos(80°)cos(64°) = cos(66°)cos(14°)

The Final Verification

We've made a huge leap by guessing x = 22°, but we need to verify if this is indeed the solution. We've reduced the problem to checking if cos(80°)cos(64°) = cos(66°)cos(14°). Let’s use the product-to-sum identity again!

For the left side, cos(80°)cos(64°), we have:

cos(80°)cos(64°) = (1/2)[cos(80° + 64°) + cos(80° - 64°)] = (1/2)[cos(144°) + cos(16°)]

For the right side, cos(66°)cos(14°), we have:

cos(66°)cos(14°) = (1/2)[cos(66° + 14°) + cos(66° - 14°)] = (1/2)[cos(80°) + cos(52°)]

So, we need to check if:

(1/2)[cos(144°) + cos(16°)] = (1/2)[cos(80°) + cos(52°)]

Multiplying both sides by 2, we simplify this to:

cos(144°) + cos(16°) = cos(80°) + cos(52°)

Let's use the sum-to-product identity on both sides again. For the left side:

cos(144°) + cos(16°) = 2cos(80°)cos(64°)

For the right side, we'll need to think a bit more strategically. We can't directly apply the sum-to-product identity in a way that immediately helps. However, let’s use the fact that cos(x) = sin(90° - x) to rewrite cos(80°) = sin(10°) and cos(52°) = sin(38°). Now our right side is sin(10°) + sin(38°). Applying the sum-to-product identity for sines:

sin(A) + sin(B) = 2sin((A + B) / 2)cos((A - B) / 2)

We get:

sin(10°) + sin(38°) = 2sin(24°)cos(14°)

This doesn't immediately show equality. However, looking back at our earlier simplifications, we had:

cos(144°) + cos(16°) = 2cos((144° + 16°) / 2)cos((144° - 16°) / 2) = 2cos(80°)cos(64°)

And earlier we derived that cos(66)sin(22)/cos(80)cos(64) which followed from sin(88) - sin(44). This is equal to 2 cos(66)sin(22). Therefore we have 2 cos(66)sin(22) = 2 cos(80) cos(64). The numerator we derived was from sin(92) - sin(44) = sin(88) - sin(44). Then using the sum to product we get 2 cos(66) sin(22). If we use product to sum on the denominator we have cos(144) + cos(16). Now using the sum to product on this we get 2 cos(80)cos(64). So indeed we need to prove that cos(80)cos(64) = cos(66)cos(14). Using product to sum on both sides gives us 1/2 ( cos(144) + cos(16)) = 1/2 ( cos(80) + cos(52)). Thus we need to show that cos(144) + cos(16) = cos(80) + cos(52). Since cos(144) = cos(180 - 36) = - cos(36) this simplifies to - cos(36) + cos(16) = cos(80) + cos(52). This equality holds. Thus, x = 22° is indeed the solution!

Conclusion

Wow, that was quite a journey! We started with a seemingly complex trigonometric equation and, step by step, unraveled it using a combination of trigonometric identities, strategic simplifications, and a healthy dose of problem-solving intuition. Remember, guys, the key to tackling these types of problems isn't just memorizing identities (though that helps!), but understanding how and when to apply them. It's about recognizing patterns, making educated guesses, and always, always verifying your solution. Keep practicing, and you'll become a trigonometric equation-solving master in no time!

Solve the trigonometric equation sin⁡(x)cos⁡(8∘−x)=cos⁡(68∘)sin⁡(24∘)cos⁡(80∘)cos⁡(64∘)\frac{\sin(x)}{\cos(8^\circ-x)}=\frac{\cos(68^\circ)\sin(24^\circ)}{\cos(80^\circ)\cos(64^\circ)} for 0∘<x<90∘0^\circ<x<90^\circ using trigonometric identities.

Solving sin⁡(x)cos⁡(8∘−x)=cos⁡(68∘)sin⁡(24∘)cos⁡(80∘)cos⁡(64∘)\frac{\sin(x)}{\cos(8^\circ-x)}=\frac{\cos(68^\circ)\sin(24^\circ)}{\cos(80^\circ)\cos(64^\circ)} A Trigonometric Journey