JEE Main 2023MathematicsTrigonometric Ratios & IdentitiesEasyMCQ

JEE Main 2023 — Trigonometric Ratios & Identities Question with Solution

JEE Main 2023 (10 Apr Shift 1)

Question

96 cosπ33 cos2π33 cos4π33 cos8π33 cos16π33 is equal to 

Choose an option

Show full solutionCorrect option: A
Correct answer
A3

Step-by-step explanation

Given,

Expression 96·cosπ33·cos2π33·cos4π33.........cos16π33

Now we know that,

cosA·cos2A·cos22A·cos23A.....·cos2n-1A=sin2nA2nsinA

Now using the above formula in given expression we get,

96·cosπ33·cos2π33·cos4π33.........cos16π33

=96×sin32π3325sinπ33

=96×sinπ-π3325sinπ33

=96×sinπ3325sinπ33 as sinπ-α=sinα

=96×132=3

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About this question

This is a previous-year question from JEE Main 2023, 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.