Math Problem Statement

Solve this

Simplify the given trigonometric expression: \( \frac{(\sin \theta + \cos \theta)^2 - 3 \sin \theta \cos \theta}{\cos \theta (\sec \theta - \csc \theta)} \cdot \frac{\sin^2 \theta - \cos^2 \theta}{\sin^3 \theta + \cos^3 \theta} \).

Solution

The equation in the image seems partially extracted due to text recognition limitations. However, from what I can make out, the given expression is as follows:

(sinθ+cosθ)23sinθcosθcosθ(secθcscθ)sin2θcos2θsin3θ+cos3θ\frac{(\sin \theta + \cos \theta)^2 - 3 \sin \theta \cos \theta}{\cos \theta (\sec \theta - \csc \theta)} \cdot \frac{\sin^2 \theta - \cos^2 \theta}{\sin^3 \theta + \cos^3 \theta}

Let me work through this expression to simplify it step by step.

  1. Expression 1 (Numerator): (sinθ+cosθ)23sinθcosθ(\sin \theta + \cos \theta)^2 - 3 \sin \theta \cos \theta Expanding the square: (sin2θ+2sinθcosθ+cos2θ)3sinθcosθ(\sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta) - 3 \sin \theta \cos \theta Since sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1, this simplifies to: 1+2sinθcosθ3sinθcosθ=1sinθcosθ1 + 2 \sin \theta \cos \theta - 3 \sin \theta \cos \theta = 1 - \sin \theta \cos \theta

  2. Expression 2 (Denominator): cosθ(secθcscθ)\cos \theta (\sec \theta - \csc \theta) We know secθ=1cosθ\sec \theta = \frac{1}{\cos \theta} and cscθ=1sinθ\csc \theta = \frac{1}{\sin \theta}, so: cosθ(1cosθ1sinθ)=1cosθsinθ=sinθcosθsinθ\cos \theta \left( \frac{1}{\cos \theta} - \frac{1}{\sin \theta} \right) = 1 - \frac{\cos \theta}{\sin \theta} = \frac{\sin \theta - \cos \theta}{\sin \theta}

  3. Expression 3 (Other factors): The second part involves: sin2θcos2θsin3θ+cos3θ\frac{\sin^2 \theta - \cos^2 \theta}{\sin^3 \theta + \cos^3 \theta} This can be rewritten as a difference of squares and cubes. Let me further simplify this next.

I'll now simplify this whole expression.The given complex expression simplifies to sinθ\sin \theta.

If you have further questions about specific steps or need more details, feel free to ask! Here are some related questions to explore:

  1. How can trigonometric identities help simplify complex expressions?
  2. What is the significance of sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 in trigonometry?
  3. How do you derive the sum and difference of cubes for trigonometric functions?
  4. What are the applications of secant (secθ\sec \theta) and cosecant (cscθ\csc \theta) identities?
  5. Can this technique be applied to other complex trigonometric expressions?

Tip: When simplifying trigonometric expressions, always look for opportunities to apply basic identities such as the Pythagorean identity.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Algebraic Manipulation
Trigonometric Identities

Formulas

Pythagorean identity: \( \sin^2 \theta + \cos^2 \theta = 1 \)
Difference of squares: \( a^2 - b^2 = (a - b)(a + b) \)
Cofunction identities: \( \sec \theta = \frac{1}{\cos \theta}, \csc \theta = \frac{1}{\sin \theta} \)

Theorems

Trigonometric Identities
Sum and Difference of Cubes

Suitable Grade Level

Grades 11-12