prove by induction, very difficultWatch

#1
Hello; i cant seem to get my head round the following question.

"prove by induction that D^n cosbx = (b^n).cos(bx+(npi)/2), where b is a constant. hence find D^n xcosbx,where D=d/dx"

0
12 years ago
#2
Do you mean D^n cosbx = (b^n).cos(bx-(npi)/2)?

Show that it's true for the first case, n=1. (The case for n=0, cosbx = cosbx, is trivial.)
d/dx cosbx = bsinbx = bcos(bx - pi/2).

Assume true for all integers up to some value n = k.
D^k cosbx = (b^k).cos(bx-(kpi)/2)
D^k+1 cosbx = d/dx [(b^k).cos(bx-(kpi)/2)]

Using the fact that cos a = sin (a - pi/2), the rest of this proof is easy.
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