Trig Question. (1 Viewer)

[currysauce]

Prove that

4cos4x cos2x cosx = cos7x + cos5x + cos3x + cosx

 

[Slidey]

4cos4x cos2x cosx = cos7x + cos5x + cos3x + cosx

LHS=4cos4xcos2xcosx
LHS=2cosx(cos6x+cos2x)
LHS=2cosxcos6x+2cos2xcosx
LHS=cos7x+cos5x+cos3x+cosx=RHS

Using the result: cosa+cosb=(cos([a+b]/2)cos([a-b]/2)/2

EDIT: I seem to have my LHS and my RHS backwards.

 

[munkaii]

4cos4x cos2x cosx = cos7x + cos5x + cos3x + cosx
RHS=2cosx(cos6x+cos2x)
RHS=2cosxcos6x+2cos2xcosx
RHS=cos7x+cos5x+cos3x+cosx=LHS

Using the result: cosa+cosb=(cos([a+b]/2)cos([a-b]/2)/2

lol seem to have your LHS and RHS backwards

 

[currysauce]

thanks ppl