Let's dive into solving the trigonometric equation ipsin(1)sesin(5)se = cos(1)secos(5)se. This might look a bit complex at first glance, but don't worry, we'll break it down step by step. We're here to make sure you understand every twist and turn of this mathematical journey.

Understanding the Basics

Before we start, let's refresh some basic trigonometric identities and concepts. Remember that trigonometric functions like sine (sin) and cosine (cos) relate angles of a right triangle to the ratios of its sides. Also, understanding the unit circle and how angles correspond to sine and cosine values is crucial.

Key Trigonometric Identities

Here are a few identities that will be helpful:

• Pythagorean Identity: sin²(x) + cos²(x) = 1
• Angle Sum and Difference Identities:
  • sin(A + B) = sin(A)cos(B) + cos(A)sin(B)
  • cos(A + B) = cos(A)cos(B) - sin(A)sin(B)

Rewriting the Equation

The given equation is: ipsin(1)sesin(5)se = cos(1)secos(5)se. It seems there might be a typo. Assuming it should be: sin(1)sin(5x) = cos(1)cos(5x), let's proceed with that.

We need to manipulate this equation to a form that we can easily solve. Our goal is to isolate the variable 'x'. Let's start by rearranging the equation.

Rearranging Terms

Move all terms to one side of the equation:

cos(1)cos(5x) - sin(1)sin(5x) = 0

Do you notice anything familiar? This form is similar to the cosine addition formula. Let's use that to simplify things further.

Applying Trigonometric Identities

Using the Cosine Addition Formula

Recall the cosine addition formula:

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

Comparing this with our equation, we can see that A = 1 and B = 5x. So, we can rewrite our equation as:

cos(1 + 5x) = 0

Now, the equation looks much simpler and easier to handle. We're on the right track, guys!

Solving for x

To solve cos(1 + 5x) = 0, we need to find the angles for which the cosine function equals zero. Remember that cosine is zero at π/2 and 3π/2 (and their coterminal angles).

So, we have:

1 + 5x = π/2 + nπ, where n is an integer.

Now, let's isolate x.

Isolating x

Subtract 1 from both sides:

5x = π/2 - 1 + nπ

Divide by 5:

x = (π/2 - 1 + nπ) / 5

Thus, the general solution for x is:

x = (π/10 - 1/5) + (nπ/5)

Where n is an integer. This means there are infinite solutions, each corresponding to a different integer value of n. Isn't that neat?

General Solutions and Specific Examples

The general solution x = (π/10 - 1/5) + (nπ/5) gives us a family of solutions. Let's look at a few specific examples to understand this better.

Examples for Different Values of n

• n = 0:

x = (π/10 - 1/5)

• n = 1:

x = (π/10 - 1/5) + (π/5) = (3π/10 - 1/5)

• n = 2:

x = (π/10 - 1/5) + (2π/5) = (5π/10 - 1/5) = (π/2 - 1/5)

And so on. Each integer value of n gives us a different solution. These solutions represent angles where the original equation holds true. Always remember that in trigonometry, we often deal with periodic functions, leading to multiple or infinite solutions.

Verification

Plugging Back into the Original Equation

To ensure our solution is correct, we can plug a few values of x back into the original equation: sin(1)sin(5x) = cos(1)cos(5x).

Let's verify with x = (π/10 - 1/5):

sin(1)sin(5(π/10 - 1/5)) = sin(1)sin(π/2 - 1)

cos(1)cos(5(π/10 - 1/5)) = cos(1)cos(π/2 - 1)

Using the identity sin(π/2 - θ) = cos(θ) and cos(π/2 - θ) = sin(θ), we get:

sin(1)cos(1 - π/2) = cos(1)sin(1 - π/2)

This confirms that our solution is correct, as both sides are equal. You can try with other values of n to further verify.

Common Mistakes to Avoid

When working on trigonometric equations, it's easy to make a few common mistakes. Being aware of these pitfalls can save you a lot of trouble.

Forgetting the General Solution

One common mistake is to only find one solution and forget that trigonometric functions are periodic. Always remember to include the general solution with '+ nπ' or '+ 2nπ', depending on the function's periodicity.

Incorrectly Applying Identities

Make sure to double-check the trigonometric identities before applying them. A small error can lead to an incorrect solution.

Calculation Errors

Be careful with algebraic manipulations, especially when dealing with fractions and negative signs. Always double-check your calculations.

Not Verifying the Solution

It's a good practice to plug your solution back into the original equation to verify that it holds true. This can help you catch any mistakes you might have made.

Alternative Approaches

While using the cosine addition formula is a straightforward approach, there are alternative ways to solve this equation. Let's explore one such method.

Using the Tangent Function

Starting from the equation sin(1)sin(5x) = cos(1)cos(5x), we can divide both sides by cos(1)cos(5x), provided that cos(1) ≠ 0 and cos(5x) ≠ 0:

(sin(1)sin(5x)) / (cos(1)cos(5x)) = 1

This simplifies to:

tan(1)tan(5x) = 1

So, tan(5x) = 1 / tan(1) = cot(1)

Since cot(1) = tan(π/2 - 1), we have:

tan(5x) = tan(π/2 - 1)

Therefore:

5x = π/2 - 1 + nπ

x = (π/2 - 1 + nπ) / 5

x = (π/10 - 1/5) + (nπ/5)

Which is the same general solution we found earlier. This alternative approach confirms our result and provides another perspective on solving the problem. How cool is that?

Conclusion

In summary, to solve the equation sin(1)sin(5x) = cos(1)cos(5x), we used trigonometric identities, particularly the cosine addition formula, to simplify the equation. We found the general solution to be x = (π/10 - 1/5) + (nπ/5), where n is an integer. We also verified our solution and discussed common mistakes to avoid.

Trigonometry can seem challenging, but with practice and a solid understanding of the basic principles, you can tackle even the most complex problems. Keep practicing, stay curious, and don't be afraid to explore different approaches. You've got this, guys!