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JEE Main20199 Jan 2019Morning ShiftMathematicsTrigonometric Ratios & IdentitiesActual

For any θ ∈ π 4 , π 2 , the expression 3 sin θ - cos θ 4 + 6 sin θ + cos θ 2 + 4 s i n 6 θ equals:

Options

  1. A13 - 4 cos 2 θ + 6 cos 4 θ
  2. B13 - 4 cos 2 θ + 6 sin 2 θ cos 2 θ
  3. C13 - 4 cos 6 θ
  4. D13 - 4 cos 4 θ + 2 sin 2 θ cos 2 θ

Correct answer

C. 13 - 4 cos 6 θ

Step-by-step solution

3 sin θ - cos θ 4 + 6 cos θ + sin θ 2 + 4 s i n 6 θ = 3 sin θ - cos θ 2 2 + 6 c o s θ + s i n θ 2 + 4 s i n 6 θ = 3 c o s 2 θ + s i n 2 θ - 2 s i n θ c o s θ 2 + 6 c o s 2 θ + s i n 2 θ + 2 s i n θ c o s θ + 4 s i n 6 θ = 3 1 - 2 s i n θ c o s θ 2 + 6 1 + 2 s i n θ c o s θ + 4 s i n 6 θ = 3 1 - 4 s i n θ c o s θ + 4 s i n 2 θ c o s 2 θ + 6 1 + 2 s i n θ c o s θ + 4 s

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