Math Problem Statement

The expression 1 + sin 2 α cos ( 2 α − 2 π ) tan ( α − 3 ( π 2 ) ) − 1 4 sin 2 α ( cot ( α 2 ) + cot ( 3 ( π 2 ) + α 2 ) ) when simplified reduces to

Solution

Let's simplify the expression step by step:

1+sin2αcos(2α2π)tan(α3(π2))14sin2α(cot(α2)+cot(3(π2)+α2))1 + \frac{\sin^2 \alpha}{\cos(2\alpha - 2\pi)} \tan\left(\alpha - 3\left(\frac{\pi}{2}\right)\right) - \frac{1}{4} \sin^2 \alpha \left(\cot\left(\frac{\alpha}{2}\right) + \cot\left(3\left(\frac{\pi}{2}\right) + \frac{\alpha}{2}\right)\right)

Step 1: Simplify the trigonometric arguments

1.1. Simplify cos(2α2π)\cos(2\alpha - 2\pi):

We know that subtracting 2π2\pi from any angle results in the same value because of periodicity: cos(2α2π)=cos(2α)\cos(2\alpha - 2\pi) = \cos(2\alpha)

1.2. Simplify tan(α3(π2))\tan\left(\alpha - 3\left(\frac{\pi}{2}\right)\right):

Shifting by 3π2\frac{3\pi}{2} corresponds to a phase shift of 270 degrees: tan(α3(π2))=tan(α+π2)\tan\left(\alpha - 3\left(\frac{\pi}{2}\right)\right) = \tan\left(\alpha + \frac{\pi}{2}\right) Using the identity tan(θ+π2)=cot(θ)\tan\left(\theta + \frac{\pi}{2}\right) = -\cot(\theta): tan(α+π2)=cot(α)\tan\left(\alpha + \frac{\pi}{2}\right) = -\cot(\alpha)

So now the expression becomes: 1+sin2αcos(2α)(cot(α))14sin2α(cot(α2)+cot(3(π2)+α2))1 + \frac{\sin^2 \alpha}{\cos(2\alpha)} \cdot (-\cot(\alpha)) - \frac{1}{4} \sin^2 \alpha \left(\cot\left(\frac{\alpha}{2}\right) + \cot\left(3\left(\frac{\pi}{2}\right) + \frac{\alpha}{2}\right)\right)

Step 2: Simplify the cotangent terms

2.1. Simplify cot(3(π2)+α2)\cot\left(3\left(\frac{\pi}{2}\right) + \frac{\alpha}{2}\right):

Using the same identity for a phase shift by 3π/23\pi/2, we get: cot(3(π2)+α2)=tan(α2)\cot\left(3\left(\frac{\pi}{2}\right) + \frac{\alpha}{2}\right) = -\tan\left(\frac{\alpha}{2}\right)

Thus, the expression simplifies to: 1sin2αcot(α)cos(2α)14sin2α(cot(α2)tan(α2))1 - \frac{\sin^2 \alpha \cot(\alpha)}{\cos(2\alpha)} - \frac{1}{4} \sin^2 \alpha \left(\cot\left(\frac{\alpha}{2}\right) - \tan\left(\frac{\alpha}{2}\right)\right)

Step 3: Further simplifications

  1. The term sin2αcot(α)\sin^2 \alpha \cot(\alpha) can be simplified using the identity cot(α)=cos(α)sin(α)\cot(\alpha) = \frac{\cos(\alpha)}{\sin(\alpha)}: sin2αcot(α)=sinαcosα\sin^2 \alpha \cot(\alpha) = \sin\alpha \cos\alpha

    Hence, the term becomes: sinαcosαcos(2α)\frac{\sin\alpha \cos\alpha}{\cos(2\alpha)}

  2. cos(2α)=2cos2(α)1\cos(2\alpha) = 2\cos^2(\alpha) - 1, which can be used if needed.

At this point, a simplified form of the original expression is: 1sinαcosαcos(2α)14sin2α(cot(α2)tan(α2))1 - \frac{\sin \alpha \cos \alpha}{\cos(2\alpha)} - \frac{1}{4} \sin^2 \alpha \left(\cot\left(\frac{\alpha}{2}\right) - \tan\left(\frac{\alpha}{2}\right)\right)

This expression doesn't immediately simplify further without additional substitutions or assumptions. Depending on specific values of α\alpha, this can be evaluated numerically.

Would you like me to continue the simplification, or do you have any specific questions about the process?


Here are 5 related questions that can expand on this:

  1. How do trigonometric identities such as angle addition and double-angle formulas help simplify expressions?
  2. What is the significance of periodicity in trigonometric functions?
  3. How does the cotangent function behave with phase shifts like π2\frac{\pi}{2} and π\pi?
  4. How can numerical methods be used to solve trigonometric expressions when exact simplification is challenging?
  5. How can trigonometric expressions like this be applied to solve problems in physics, particularly in wave mechanics?

Tip: When simplifying trigonometric expressions, always check for opportunities to apply standard identities, such as double angle or sum formulas, to make the expression more manageable.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Periodic Functions
Trigonometric Identities

Formulas

cos(θ − 2π) = cos(θ)
tan(θ + π/2) = -cot(θ)
cot(θ) = cos(θ) / sin(θ)
cos(2θ) = 2cos^2(θ) - 1

Theorems

Trigonometric Angle Addition
Double-Angle Identities
Periodicity of Trigonometric Functions

Suitable Grade Level

College-level (Advanced Trigonometry)