induction. (1 Viewer)

[conics2008]

Use Mathematical Induction to show that cos(x + piN) = (-1)^n cosx for all positive integers n >= 1

i got up to

-cos(x+kpi) +(-1)^k cos(x) ??

 

[midifile]

Use Mathematical Induction to show that cos(x + pi) = (-1)^n cosx for all positive integers n >= 1

should there be an n on the lhs?

 

[doink]

for n=k rearrange to make (-1)^k = cosx (x+ pi) / cos x and use it for n=k+1 making (-1)^k+1 = (-1)^k(-1).

Not sure if it will work but thats as far as i can go in my head only.

 

[conics2008]

yes there is an n in the cos(x+pin)

 

[lolokay]

Use Mathematical Induction to show that cos(x + piN) = (-1)^n cosx for all positive integers n >= 1

i got up to

-cos(x+kpi) +(-1)^k cos(x) ??

cos(x + n.pi + pi)
= cos(x + n.pi)cospi + sin(x+n.pi)sin pi
= -cos(x+n.pi)

(-1)^(n+1) cosx
= -(-1)^n cos x

-cos(x+n.pi) = -(-1)^n cos x
cos(x+n.pi) = (-1)^n cos x

so if true for n, then true for n+1
if n = 1
cos(x+pi)
= cosx cospi - sinx sinpi
= -1cosx
= RHS

so true for all n>= 1

 

[3unitz]

assume true for n = k
cos(x + k*pi) = (-1)^k cos(x)

n = k+1:
LHS = cos[x + (k+1)*pi], RHS = (-1)^(k+1) cos (x)

LHS = cos [(x + k*pi) + pi]
= cos(x + k*pi) cos(pi) - sin(x + k*pi) sin(pi)
= - cos(x + k*pi)
= - (-1)^k cos(x) using assumption
= (-1)^(k+1) cos (x)
= RHS

 

[midifile]

Prove true for n=1
I can tbe bothered to do (im sure you can do that)

Assume true for n=k
cos(x+kpi)=(-1)^kcosx

Prove true for n=k+1
rtp: cos(x+(k+1)pi) = (-1)^(k+1)cosx
lhs= cos(x+kpi+pi)
= -cos(x+kpi)
= -(-1)^kcosx
= (-1)^k(-1)^1cosx
= (-1)^(k+1)cosx
=rhs
Therefore true for n=k+1 blah blah blah

 

[conics2008]

ok thanks i found out where i went wrong.. xD thanks all =]