f(x)=cos(3pix-(pi/2))
Find the set on which the function is continuous (in all cases, the domain). Do so analytically, identify the type of function and the appropriate algebraic rule for determining domain.
Answers
This function is continuous on all R, it is simply a transformation of cos(x) to cos(ax+b), where a=3pi and b=-pi/2. cos(x) is continuous on all R, so it is it's transformed version. If you want the proof, for each t in R, lim (x->t-) f(x) = lim (x->t+) f(x) = cos(3pi*t-pi/2)
30 January 2012
the proof is similar to the proof that cos(x) is continuous everywhere.
30 January 2012
just put the usual ε
30 January 2012
cos(t+δ)-cos(t)
30 January 2012
cos(t+δ)=cos(t)cos(δ)-sin(t)sin(δ)
30 January 2012
cos(t+δ)-cos(t)=cos(t)(cos(δ)-1)-sin(t)sin(δ)
30 January 2012
now, if δ is small, so is sin(δ)
30 January 2012
and for δ>0 and δ small, cos(δ)=(1-sin(δ)^2)^(1/2)
30 January 2012
just put α=sin(δ)
30 January 2012
|cos(t+δ)-cos(t)|
30 January 2012
(1-cos(δ))+sin(δ)=1-(1-α^2)^(1/2)+α
30 January 2012
1-(1-α^2)^(1/2)+α
30 January 2012
(1+α-ε )
30 January 2012
etc...
30 January 2012
oops sorry, let's take it back from |cos(t+δ)-cos(t)|=|cos(t)(cos(δ)-1)-sin(t)sin(δ)|
30 January 2012
|cos(t)|
30 January 2012
now |cos(δ)-1|=|1-cos(δ)|
30 January 2012
|cos(δ)-1|+|sin(δ)|
30 January 2012
|cos(δ)-1|=|sin(δ)|^2 / |1+cos(δ)|
30 January 2012
now, |δ| is very small, so cos(δ)>0, therefore |sin(δ)|^2 / |1+cos(δ)|
30 January 2012
and since |sin(δ)|
30 January 2012
therefore |cos(δ)-1|+|sin(δ)|
30 January 2012
|sin(δ)|
30 January 2012
|δ|
30 January 2012
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