Math Problem Statement

Prove \( \sin^2(\theta) = \frac{1}{2} - \frac{1}{2}\cos(2\theta) \) using Euler's equations.

Solution

To prove that sin2(θ)=1212cos(2θ)\sin^2(\theta) = \frac{1}{2} - \frac{1}{2}\cos(2\theta) using Euler's equations, let's proceed step by step.

Euler's equations:

eiθ=cos(θ)+isin(θ),eiθ=cos(θ)isin(θ).e^{i\theta} = \cos(\theta) + i\sin(\theta), \quad e^{-i\theta} = \cos(\theta) - i\sin(\theta).

From these, we can derive expressions for sin(θ)\sin(\theta) and cos(θ)\cos(\theta): cos(θ)=eiθ+eiθ2,sin(θ)=eiθeiθ2i.\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}, \quad \sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}.

Step 1: Express sin2(θ)\sin^2(\theta) using Euler's equations

sin2(θ)=(eiθeiθ2i)2.\sin^2(\theta) = \left( \frac{e^{i\theta} - e^{-i\theta}}{2i} \right)^2.

Expanding this: sin2(θ)=(eiθeiθ)24.\sin^2(\theta) = \frac{(e^{i\theta} - e^{-i\theta})^2}{-4}.

Simplify the numerator: (eiθeiθ)2=e2iθ2+e2iθ.(e^{i\theta} - e^{-i\theta})^2 = e^{2i\theta} - 2 + e^{-2i\theta}.

Thus: sin2(θ)=e2iθ2+e2iθ4.\sin^2(\theta) = \frac{e^{2i\theta} - 2 + e^{-2i\theta}}{-4}.

Step 2: Separate the terms

Split into real and imaginary components: sin2(θ)=14(e2iθ+e2iθ2).\sin^2(\theta) = -\frac{1}{4} \left( e^{2i\theta} + e^{-2i\theta} - 2 \right).

Combine the exponential terms: e2iθ+e2iθ=2cos(2θ).e^{2i\theta} + e^{-2i\theta} = 2\cos(2\theta).

Thus: sin2(θ)=14(2cos(2θ)2).\sin^2(\theta) = -\frac{1}{4} \left( 2\cos(2\theta) - 2 \right).

Step 3: Simplify the expression

Factor out 22: sin2(θ)=142(cos(2θ)1).\sin^2(\theta) = -\frac{1}{4} \cdot 2 \left( \cos(2\theta) - 1 \right).

Simplify further: sin2(θ)=1212cos(2θ).\sin^2(\theta) = \frac{1}{2} - \frac{1}{2}\cos(2\theta).

Final Result:

sin2(θ)=1212cos(2θ),\sin^2(\theta) = \frac{1}{2} - \frac{1}{2}\cos(2\theta), as required.


Would you like further clarification or additional examples? Here are some related questions to explore:

  1. Can you prove cos2(θ)=12+12cos(2θ)\cos^2(\theta) = \frac{1}{2} + \frac{1}{2}\cos(2\theta) using the same approach?
  2. How are Euler's formulas used to derive trigonometric identities generally?
  3. Why does cos(2θ)\cos(2\theta) appear in this result?
  4. How does this identity relate to the double-angle formulas for sine and cosine?
  5. Can this proof be adapted for other trigonometric powers like sin4(θ)\sin^4(\theta)?

Tip: Understanding the connection between exponential and trigonometric functions simplifies complex proofs significantly!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometry
Complex Numbers

Formulas

Euler's formula: \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \)
\( \cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2} \)
\( \sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i} \)

Theorems

Euler's Formula

Suitable Grade Level

Undergraduate Math or Advanced High School