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`        Find the solution for Nth derivative of cosx*cos3x*cos5x`
one year ago

Arun
23047 Points
```							cosx cos2x cos3x = 1/2 cos2x [ 2 cosx cos3x] [ cosC + cosD = 2 cos(C+D)/2 cos(C- D)/2 ] =1/2 cos2x [ cos4x + cos2x] = 1/4 [ 2 cos2x cos4x + 2 cos^22x ] = 1/4 [ cos6x + cos 2x + ( 1+ cos4x)] [ as cos2x = 2 cos^2x - 1 ] =1/4 cos6x +1/4 cos2x + 1/4 + 1/4 cos4x  now D^n cos(ax+ b) = a^n cos( ax+ b+ npi/2) so D^n [cosx cos2x cos3x = 1/4D^n[ cos2x+ cos4x+ cos6x ] [ d/dx 1= 0 ] = 1/4[ 2^ncos(2x+npi/2)+4^n cos(4x+npi/2) + 6^n cos(6x+ npi/2) ]
```
one year ago
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### Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions