Math for fun, sin(sin(z))=1 | Summary and Q&A

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February 2, 2021
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blackpenredpen
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Math for fun, sin(sin(z))=1

TL;DR

Solving the equation sin(sin(x)) = 1 requires complex number solutions and involves infinitely many equations with infinitely many complex solutions.

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Key Insights

  • ☺️ The equation sin(sin(x)) = 1 requires finding values of x that satisfy sin(x) = pi/2 on the unit circle.
  • ❓ When solving the generalized equation sin(sin(z)) = 1, there are infinitely many equations and infinitely many complex solutions.
  • ❓ The quadratic equation derived from sin(sin(z)) = 1 can be solved using the quadratic formula to obtain two solutions for e^iz.
  • 🤪 The final solution for z in sin(sin(z)) = 1 involves complex exponentials, natural logarithm, and integer parameters.
  • 🤪 The importance of using consistent variable names (e.g., using "z" for complex numbers) is emphasized.
  • #️⃣ The equation involves trigonometric functions, complex numbers, and exponential functions.
  • 🤪 The concept of complex solutions is introduced, highlighting the need to consider the domain of z as integers.

Transcript

okay let's do something for fun we have a very interesting question for you guys we are going to solve sine of sine of x is equal to one it doesn't look so bad huh but it is not only it's bad but it's also complex you'll see anyway though let's see first of all of course we have to look at sine of what will be one and of course just refer to the un... Read More

Questions & Answers

Q: How can the equation sin(sin(x)) = 1 be solved?

To solve sin(sin(x)) = 1, we need to find the values of x that satisfy sin(x) = pi/2. These values can be derived by considering the unit circle and the trigonometric function values.

Q: What happens when solving the generalized equation sin(sin(z)) = 1?

When solving sin(sin(z)) = 1 in the complex number domain, we encounter infinitely many equations due to the presence of an integer, n, which leads to infinitely many complex solutions for z.

Q: How is the equation sin(sin(z)) = 1 simplified using the complex exponential form?

By representing sin(z) using the complex exponential form (e^iz - e^-iz)/2i, the equation sin(sin(z)) = 1 can be simplified to e^iz - e^-iz = 2i.

Q: What is the quadratic equation derived from sin(sin(z)) = 1?

After simplifying, sin(sin(z)) = 1 leads to a quadratic equation in terms of e^iz, which can be solved using the quadratic formula resulting in two possible solutions for e^iz.

Summary & Key Takeaways

  • The equation sin(sin(x)) = 1 can be solved by finding the values of x that satisfy sin(x) = pi/2.

  • To solve for complex number solutions, the equation is generalized to sin(sin(z)) = 1, where z represents a complex variable.

  • Solving sin(sin(z)) = 1 leads to infinitely many equations and infinitely many complex solutions.

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