Solve: Arcsin(5/13) + Arcsin(7/25) = Arccos(253/325)

by Jhon Lennon 53 views

Let's dive into solving this trigonometric equation! It looks a bit intimidating at first, but we'll break it down step by step. The goal here is to prove that arcsin(5/13) + arcsin(7/25) equals arccos(253/325). We'll use trigonometric identities to simplify and verify this equation. So, buckle up, guys, it's gonna be a fun ride!

Step 1: Understanding the Basics

First, let's clarify what arcsin and arccos mean. Arcsin(x) gives you the angle whose sine is x, and arccos(x) gives you the angle whose cosine is x. In other words, if sin(θ) = x, then arcsin(x) = θ, and if cos(θ) = x, then arccos(x) = θ. These are inverse trigonometric functions, and understanding them is crucial for solving this problem. Remember, the range of arcsin is [-π/2, π/2], and the range of arccos is [0, π].

Now, let's define:

A = arcsin(5/13) B = arcsin(7/25) C = arccos(253/325)

Our mission is to prove that A + B = C.

Step 2: Using Trigonometric Identities

To proceed, we'll use the sine addition formula: sin(A + B) = sin(A)cos(B) + cos(A)sin(B)

We know sin(A) = 5/13 and sin(B) = 7/25. We need to find cos(A) and cos(B).

Using the Pythagorean identity sin^2(A) + cos^2(A) = 1: cos^2(A) = 1 - sin^2(A) = 1 - (5/13)^2 = 1 - 25/169 = 144/169 So, cos(A) = √(144/169) = 12/13 (since A is in the range of arcsin, cosine is positive).

Similarly, for cos(B): cos^2(B) = 1 - sin^2(B) = 1 - (7/25)^2 = 1 - 49/625 = 576/625 So, cos(B) = √(576/625) = 24/25 (since B is in the range of arcsin, cosine is positive).

Step 3: Calculating sin(A + B)

Now we can find sin(A + B): sin(A + B) = sin(A)cos(B) + cos(A)sin(B) = (5/13)(24/25) + (12/13)(7/25) = (120/325) + (84/325) = 204/325

So, sin(A + B) = 204/325.

Step 4: Finding cos(C)

We know that C = arccos(253/325), which means cos(C) = 253/325. Now we need to find sin(C) to compare it with sin(A+B).

Using the Pythagorean identity sin^2(C) + cos^2(C) = 1: sin^2(C) = 1 - cos^2(C) = 1 - (253/325)^2 = 1 - 64009/105625 = 41616/105625 So, sin(C) = √(41616/105625) = 204/325 (since C is in the range of arccos, sine is positive).

Step 5: Verifying A + B = C

We have found that: sin(A + B) = 204/325 sin(C) = 204/325

Since sin(A + B) = sin(C), it implies that A + B = C or A + B = π - C. However, we must consider the range of the angles to determine the correct relationship.

A = arcsin(5/13) is approximately 22.62 degrees. B = arcsin(7/25) is approximately 16.26 degrees. So, A + B is approximately 38.88 degrees.

C = arccos(253/325) is approximately 38.88 degrees.

Since A + B and C are both in the first quadrant (0 to 90 degrees), A + B = C.

Thus, arcsin(5/13) + arcsin(7/25) = arccos(253/325) is verified.

Conclusion

Alright, we've successfully shown that arcsin(5/13) + arcsin(7/25) = arccos(253/325). By using trigonometric identities and carefully considering the ranges of the inverse trigonometric functions, we were able to verify the equation. Remember, guys, practice makes perfect, so keep exploring those trig functions!

Delving deeper into inverse trigonometric functions, it's super useful to understand their properties and how they interact with each other. These functions, namely arcsin(x), arccos(x), and arctan(x), are the inverses of the basic trigonometric functions sine, cosine, and tangent. They help us find angles when we know the ratio of sides in a right triangle.

Understanding Arcsin(x)

The arcsin(x) function, also written as sin⁻¹(x), returns the angle whose sine is x. The domain of arcsin(x) is [-1, 1], and its range is [-π/2, π/2]. This means that the output angle will always be between -90 degrees and 90 degrees. When you're solving equations involving arcsin, it's crucial to remember this range to ensure your solutions are valid.

For example, if you have sin(θ) = 0.5, then arcsin(0.5) = π/6 (which is 30 degrees). This tells you the angle θ for which the sine is 0.5.

Understanding Arccos(x)

The arccos(x) function, written as cos⁻¹(x), returns the angle whose cosine is x. Like arcsin(x), the domain of arccos(x) is also [-1, 1], but its range is [0, π]. So, the output angle will always be between 0 degrees and 180 degrees. This difference in range is significant when dealing with trigonometric equations because cosine is positive in both the first and fourth quadrants.

For instance, if cos(θ) = 0.5, then arccos(0.5) = π/3 (which is 60 degrees). Keep in mind that while -π/3 also has a cosine of 0.5, it falls outside the range of arccos.

Understanding Arctan(x)

The arctan(x) function, written as tan⁻¹(x), gives you the angle whose tangent is x. The domain of arctan(x) is all real numbers, and its range is (-π/2, π/2). This means the output angle will be between -90 degrees and 90 degrees. Arctan is particularly useful because the tangent function can take any real number as a ratio of sides.

For example, if tan(θ) = 1, then arctan(1) = π/4 (which is 45 degrees). Unlike sine and cosine, tangent doesn't have a restricted domain, making arctan versatile for various problems.

Practical Applications and Tips

When solving complex trigonometric equations, remember these tips:

  1. Always check the domain and range: Make sure your solutions fall within the valid ranges of the inverse trigonometric functions you're using.
  2. Use trigonometric identities: Identities like sin²(x) + cos²(x) = 1, tan(x) = sin(x)/cos(x), and the angle sum/difference formulas can simplify equations.
  3. Draw diagrams: Visualizing the problem can often help you understand the relationships between angles and sides.
  4. Be mindful of multiple solutions: Trigonometric functions are periodic, so there can be multiple angles that satisfy a given condition. Make sure you find all possible solutions within the specified range.

By mastering these functions and techniques, you'll be well-equipped to tackle a wide range of trigonometric problems. Keep practicing, and you'll become a trig whiz in no time, guys!

Let's kick things up a notch and explore some advanced trigonometric techniques that can be incredibly useful when dealing with complex problems. These techniques include using half-angle formulas, product-to-sum formulas, and understanding how to manipulate trigonometric equations to find elegant solutions.

Half-Angle Formulas

Half-angle formulas are powerful tools that allow you to find the trigonometric functions of half an angle when you know the trigonometric functions of the full angle. These formulas are derived from the double-angle formulas and are extremely useful in simplifying expressions and solving equations.

The half-angle formulas are as follows:

sin(x/2) = ±√((1 - cos(x))/2) cos(x/2) = ±√((1 + cos(x))/2) tan(x/2) = ±√((1 - cos(x))/(1 + cos(x))) = sin(x)/(1 + cos(x)) = (1 - cos(x))/sin(x)

The ± sign indicates that you need to determine the correct sign based on the quadrant in which x/2 lies. For example, if x/2 is in the first or second quadrant, sin(x/2) will be positive. If it's in the third or fourth quadrant, sin(x/2) will be negative. The same logic applies to cosine and tangent.

Example

Suppose you want to find sin(15°). You know that 15° is half of 30°, and you know the values of sin(30°) and cos(30°).

sin(30°) = 1/2 cos(30°) = √3/2

Using the half-angle formula for sine: sin(15°) = sin(30°/2) = √((1 - cos(30°))/2) = √((1 - √3/2)/2) = √((2 - √3)/4) = √(2 - √3)/2

Product-to-Sum Formulas

Product-to-sum formulas allow you to convert products of trigonometric functions into sums or differences, which can simplify complex expressions and make them easier to evaluate.

The product-to-sum formulas are:

sin(A)cos(B) = 1/2[sin(A + B) + sin(A - B)] cos(A)sin(B) = 1/2[sin(A + B) - sin(A - B)] cos(A)cos(B) = 1/2[cos(A + B) + cos(A - B)] sin(A)sin(B) = 1/2[cos(A - B) - cos(A + B)]

These formulas are derived from the angle sum and difference formulas and are invaluable for simplifying expressions involving products of sines and cosines.

Example

Simplify sin(3x)cos(x) using the product-to-sum formula: sin(3x)cos(x) = 1/2[sin(3x + x) + sin(3x - x)] = 1/2[sin(4x) + sin(2x)]

Manipulating Trigonometric Equations

One of the most important skills in advanced trigonometry is the ability to manipulate trigonometric equations to find elegant solutions. This often involves using a combination of trigonometric identities, algebraic techniques, and creative problem-solving.

Tips for Manipulating Equations

  1. Look for Opportunities to Use Identities: Always be on the lookout for ways to apply trigonometric identities to simplify the equation. This might involve rewriting terms using Pythagorean identities, angle sum/difference formulas, or half-angle formulas.
  2. Factor and Simplify: Factoring can often help you break down a complex equation into simpler parts. Look for common factors and use algebraic techniques to simplify the equation.
  3. Isolate Trigonometric Functions: Try to isolate trigonometric functions on one side of the equation. This can make it easier to solve for the unknown angle.
  4. Check for Extraneous Solutions: When solving trigonometric equations, it's important to check for extraneous solutions. These are solutions that satisfy the simplified equation but not the original equation. Always plug your solutions back into the original equation to make sure they are valid.

Example

Solve the equation sin(2x) = cos(x) for x in the interval [0, 2π]:

sin(2x) = 2sin(x)cos(x) So, the equation becomes: 2sin(x)cos(x) = cos(x) 2sin(x)cos(x) - cos(x) = 0 cos(x)(2sin(x) - 1) = 0

This gives us two possibilities:

  1. cos(x) = 0, which means x = π/2 or x = 3π/2
  2. 2sin(x) - 1 = 0, which means sin(x) = 1/2, so x = π/6 or x = 5π/6

Thus, the solutions in the interval [0, 2π] are x = π/6, π/2, 5π/6, and 3π/2.

By mastering these advanced trigonometric techniques, you'll be able to tackle even the most challenging problems with confidence. Keep practicing, and you'll become a true trigonometry expert, guys! Remember, practice makes perfect, and with a solid understanding of these concepts, you'll be well on your way to mastering trigonometry.