the sine triangle problem
TLDRIn this educational video, the presenter explores the intriguing concept of a right triangle with sides defined by trigonometric functions: 2x, sin(2x), and a hypotenuse of sin(3x). The video challenges viewers to apply the Pythagorean theorem to determine the values of x that make the triangle legitimate. Through a step-by-step process, the presenter uses trigonometric identities and algebraic manipulation to solve for x, ultimately finding that x = Ο/6 + 2nΟ, where n is an integer. The solution is verified by substituting the value back into the original equations, confirming the validity of the right triangle. The video concludes by highlighting the mathematical beauty and satisfaction derived from solving such problems.
Takeaways
- π The presenter introduces a right triangle with sides labeled as 2x, 3x, and the hypotenuse as \( \sin(3x) \).
- π The Pythagorean theorem is used to set up the equation for the sides of the triangle: \( \sin^2(x) + (\sin(2x))^2 = (\sin(3x))^2 \).
- π The double angle identity for sine is applied: \( \sin(2x) = 2\sin(x)\cos(x) \).
- 𧩠The triple angle identity for sine is used, resulting in \( 3\sin(x) - 4\sin^3(x) \).
- π’ The equation is expanded and simplified to find a relationship between \( \sin(x) \) and \( \sin(2x) \).
- βοΈ Factoring is used extensively to simplify the equation into a quadratic form in terms of \( \sin^2(x) \).
- π The difference of squares is applied to further simplify the equation: \( (2\sin(x) - 1)(2\sin(x) + 1) = 0 \).
- π« The presenter eliminates solutions that do not satisfy the conditions of a right triangle (positive side lengths).
- π The only valid solution for \( \sin(x) \) is found to be \( \sin(x) = \frac{1}{2} \), leading to \( x = \frac{\pi}{6} + 2n\pi \) where \( n \) is an integer.
- π The presenter concludes by verifying the solution by substituting \( x = \frac{\pi}{6} \) back into the original triangle sides and confirming it forms a right triangle.
Q & A
What is the shape of the triangle discussed in the transcript?
-The shape of the triangle discussed is a right triangle with sides denoted as S of X, S of 2X, and the hypotenuse being S of 3X.
What theorem is used to set up the equation for the right triangle?
-The Pythagorean theorem is used to set up the equation for the right triangle, equating the sum of the squares of the two shorter sides to the square of the hypotenuse.
What is the double angle identity used in the script?
-The double angle identity used is sin(2X) = 2 * sin(X) * cos(X).
What is the triple angle identity mentioned in the script?
-The script refers to a triple angle identity, but it is not explicitly given. However, it is implied that it involves sin(3X) in some form.
How does the script approach solving for X?
-The script approaches solving for X by setting up an equation using trigonometric identities and then factoring and simplifying to find the values of X that satisfy the condition of forming a legitimate right triangle.
What are the conditions for the sides of the triangle to be valid?
-The conditions for the sides of the triangle to be valid are that all sides must be positive numbers, with no zero, negative, or imaginary values.
What is the first solution for X that the script identifies?
-The first solution for X that the script identifies is X = Ο/6.
How does the script generalize the solution for X?
-The script generalizes the solution for X to be X = Ο/6 + 2nΟ, where n is an integer.
What is the significance of squaring the trigonometric functions in the script?
-Squaring the trigonometric functions is significant because it allows for the simplification of the equation and helps in finding the values of X that satisfy the condition of forming a right triangle, regardless of the sign of the sine function.
Why does the script discard certain solutions for X?
-The script discards certain solutions for X because they do not result in all sides of the triangle being positive, which is a requirement for a legitimate right triangle.
How does the script verify that the solution for X is correct?
-The script verifies that the solution for X is correct by plugging the value of X back into the original trigonometric expressions and showing that the equation holds true, confirming that the sides of the triangle satisfy the Pythagorean theorem.
Outlines
πΊ Understanding the Sign Triangle
This paragraph explains the setup of a right triangle with sides \( \sin x \), \( \sin 2x \), and \( \sin 3x \). The speaker walks through the process of determining for which \( x \) values this setup forms a valid right triangle, utilizing the Pythagorean theorem. By squaring the sides and applying trigonometric identities, the equation \( \sin^2 x + \sin^2 2x = \sin^2 3x \) is formed. This is then expanded and simplified, demonstrating the methodical approach to solving this trigonometric problem.
π§ Ensuring Valid Triangle Sides
This paragraph focuses on ensuring the sides of the triangle are positive and valid. The speaker evaluates different \( \sin x \) values, ruling out zero, negative, and invalid values. The correct solution is identified as \( 2 \sin x - 1 = 0 \), leading to \( \sin x = \frac{1}{2} \) and \( x = \frac{\pi}{6} \). Other potential values are eliminated based on the requirement for all triangle sides to be positive.
π Verifying the Solution
This paragraph verifies the solution by substituting \( x = \frac{\pi}{6} \) into the original equation. It demonstrates that the sides of the triangle indeed satisfy the Pythagorean theorem. By calculating \( \sin \left( \frac{\pi}{6} \right)^2 \), \( \sin \left( \frac{\pi}{3} \right)^2 \), and \( \sin \left( \frac{\pi}{2} \right)^2 \), the equality is confirmed, ensuring the setup is correct. The paragraph concludes with an acknowledgment of the interesting and surprising nature of the problem.
Mindmap
Keywords
π‘Right Triangle
π‘Pythagorean Theorem
π‘Sine Function
π‘Double Angle Identity
π‘Triple Angle Identity
π‘Quadratic Equation
π‘Factoring
π‘Special Right Triangles
π‘Reference Triangle Method
π‘Positive Sides
π‘Integer
Highlights
Introduction of a right triangle with sides of 2x, sine of 2x, and the hypotenuse being sine of 3x.
Use of the Pythagorean theorem to set up the equation for the right triangle.
Application of the double angle identity for sine (2 * sin x * cos x).
Introduction of a triple angle identity for sine.
Equation simplification leading to a quadratic form in terms of sine squared.
Expansion and simplification of terms to form a solvable equation.
Combining like terms to isolate sine to the sixth power.
Factoring out common terms to simplify the equation further.
Identification of potential solutions by setting factors equal to zero.
Exclusion of solutions that do not satisfy the conditions of a right triangle.
Final consideration of valid solutions for sine x leading to a legitimate right triangle.
Use of the reference triangle method to find specific angle solutions.
Verification of the solution by plugging in values into the original equations.
Conclusion that x equals pi/6 + 2nPi (where n is an integer) is the only valid solution.
Demonstration of the solution's validity through substitution and calculation.
Emphasis on the importance of ensuring all sides of the triangle are positive.
Transcripts
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