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Question

The value of 2 cosx - cos 3x - cos 5x - 16 cos3x sin2x is
(a) 2
(b) 1
(c) 0
(d) −1

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Solution

(c) 0

We have,2cosx-cos3x-cos5x-16cos3x sin2x=2cosx-cos3x-cos5x-16cos3x+3cosx4×1-cos2x2=2cosx-cos3x-cos5x-2cos3x+3cosx1-cos2x=2cosx-cos3x-cos5x-2cos3x-cos3x cos2x+3cosx-3cosx cos2x=2cosx-cos3x-cos5x-2cos3x+3cosx+2cos3x cos2x+32cosx cos2x=2cosx-cos3x-cos5x-2cos3x+3cosx+cos5x+cosx+3cos3x+3cosx 2cosAcosB=cosA+B+cosA-B=2cosx-cos3x-cos5x-2cos3x-6cosx+cos5x+cosx+3cos3x+3cosx=0

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