Simplify the expression: [(cos 3θ + 2cos 5θ + cos 7θ) / (cos θ + 2cos 3θ + cos 5θ)] + (sin 2θ * tan 3θ).

Introduction / Context:This is a structured trigonometric simplification problem. The numerator and denominator are symmetric in cosine terms (3θ, 5θ, 7θ and θ, 3θ, 5θ), suggesting sum-to-product grouping. After simplifying the fraction, the remaining sin 2θ * tan 3θ term neatly cancels a part of the cosine expansion, leaving a very simple final result.

Given Data / Assumptions:

• Expression: [(cos 3θ + 2cos 5θ + cos 7θ)/(cos θ + 2cos 3θ + cos 5θ)] + sin 2θ * tan 3θ
• Use: cos X + cos Y = 2*cos((X + Y)/2)*cos((X - Y)/2)
• Use: tan 3θ = sin 3θ / cos 3θ
• Use: cos(3θ + 2θ) = cos 3θ*cos 2θ - sin 3θ*sin 2θ

Concept / Approach:Group paired cosine terms (cos 3θ + cos 7θ) and (cos θ + cos 5θ) to factor out common multipliers. Reduce the fraction to cos 5θ / cos 3θ, then combine with sin 2θ*sin 3θ / cos 3θ and apply the cosine addition identity to collapse everything to cos 2θ.

Step-by-Step Solution:

Verification / Alternative check:Pick θ = 0: the expression becomes (1 + 2*1 + 1)/(1 + 2*1 + 1) + 0 = 1, and cos 0 = 1, so it matches.

Why Other Options Are Wrong:

Common Pitfalls:Not grouping cosine pairs correctly, forgetting tan 3θ = sin 3θ / cos 3θ, or missing the cancellation of sin 2θ*sin 3θ terms.

Final Answer:cos 2θ