JEE Main 2019MathematicsTrigonometric Ratios & IdentitiesMediumMCQ

JEE Main 2019 — Trigonometric Ratios & Identities Question with Solution

JEE Main 2019 (09 Jan Shift 1)

Question

For any θ∈π4,π2, the expression 3sinθ-cosθ4+6sinθ+cosθ2+4 sin6θ equals:

Choose an option

Show full solutionCorrect option: C
Correct answer
C13-4cos6θ

Step-by-step explanation

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

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About this question

This is a previous-year question from JEE Main 2019, covering the Trigonometric Ratios & Identities chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.