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)
advmc
30 January 2012
the proof is similar to the proof that cos(x) is continuous everywhere.
advmc
30 January 2012
just put the usual ε
advmc
30 January 2012
cos(t+δ)-cos(t)
advmc
30 January 2012
cos(t+δ)=cos(t)cos(δ)-sin(t)sin(δ)
advmc
30 January 2012
cos(t+δ)-cos(t)=cos(t)(cos(δ)-1)-sin(t)sin(δ)
advmc
30 January 2012
now, if δ is small, so is sin(δ)
advmc
30 January 2012
and for δ>0 and δ small, cos(δ)=(1-sin(δ)^2)^(1/2)
advmc
30 January 2012
just put α=sin(δ)
advmc
30 January 2012
|cos(t+δ)-cos(t)|
advmc
30 January 2012
(1-cos(δ))+sin(δ)=1-(1-α^2)^(1/2)+α
advmc
30 January 2012
1-(1-α^2)^(1/2)+α
advmc
30 January 2012
(1+α-ε )
advmc
30 January 2012
etc...
advmc
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)|
advmc
30 January 2012
|cos(t)|
advmc
30 January 2012
now |cos(δ)-1|=|1-cos(δ)|
advmc
30 January 2012
|cos(δ)-1|+|sin(δ)|
advmc
30 January 2012
|cos(δ)-1|=|sin(δ)|^2 / |1+cos(δ)|
advmc
30 January 2012
now, |δ| is very small, so cos(δ)>0, therefore |sin(δ)|^2 / |1+cos(δ)|
advmc
30 January 2012
and since |sin(δ)|
advmc
30 January 2012
therefore |cos(δ)-1|+|sin(δ)|
advmc
30 January 2012
|sin(δ)|
advmc
30 January 2012
|δ|
advmc
30 January 2012
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