For any θ ∈ π 4 , π 2 , the expression 3 sin θ - cos θ 4 + 6 sin θ + cos…

For any θπ4,π2, the expression 3sinθ-cosθ4+6sinθ+cosθ2+4 sin6θ equals:
  1. 13-4cos2θ+6cos4θ
  2. 13-4cos2θ+6sin2θcos2θ
  3. 13-4cos6θ
  4. 13-4cos4θ+2sin2θcos2θ

Solution

3sinθ-cosθ4+6cosθ+sinθ2+4sin6θ

=3sinθ-cosθ22+6cosθ+sinθ2+4sin6θ

=3cos2θ+sin2θ-2sinθcosθ2+6cos2θ+sin2θ+2sinθcosθ+4sin6θ

=31-2sinθcosθ2+61+2sinθcosθ+4sin6θ

=31-4sinθcosθ+4sin2θcos2θ+61+2sinθcosθ+4sin6θ

=9+12sin2θcos2θ+4sin6θ

=9+121-cos2θcos2θ+41-cos2θ3

=9+12cos2θ-12cos4θ+41-3cos2θ+3cos4θ-cos6θ

=13-4cos6θ

Asked in: JEE Main 2019 (09 Jan Shift 1)

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