induction (1 Viewer)

[ezzy85]

theres an induction q i saw recently:

sin(npi+x) = (-1)ⁿsinx

Ive got it out except for the last line. im up to:

when n = k+1

sin((k+1)pi + x) = sinx cos(-kpi)

all ive got left is to show is that cos(-kpi) = (-1)^k+1

 

[spice girl]

Originally posted by ezzy85
theres an induction q i saw recently:

sin(npi+x) = (-1)ⁿsinx

Ive got it out except for the last line. im up to:

when n = k+1

sin((k+1)pi + x) = sinx cos(-kpi)

all ive got left is to show is that cos(-kpi) = (-1)^k+1

Trick here is to do it for n being odd, and for n being even, separately.

n even:
n = 0 obviously true
assume n=k
then sin(kpi+x) = (-1)^ksinx = sinx (as (-1)^k = 1)
sin({k+2}pi + x) = sin(kpi + x + 2pi) = sin(kpi + x) = sinx = (-1)^k+2sinx

n being odd, do practically the same thing

 

[ezzy85]

cool, thanks

 

[spice girl]

This is from noticing that sin has a period of 2pi, and sin(x+pi) = -sinx

 

[KeypadSDM]

How about this methodology:
assume true for n = k
.: Sin[kPi + x] = (-1)^k * Sin[x]

For n = k + 1
Sin[(k+ 1)Pi + x]
= Sin[(kPi + x) + Pi]
= Sin[kPi + x]Cos[Pi] + Sin[Pi]Cos[kPi + x]
= -1 * Sin[kPi + x]
= -1 * (-1)^k * Sin[x]
= (-1)^(k + 1) * Sin[x]
.: true for n = k + 1 if true for n = k
.: true for all integral k >= 0 (or whatever one you started at...)

this one's a lot quicker because u don't have to do both odd and evens...

 

[ezzy85]

that one also looks good. Thanks for that.