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wany
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Homework Statement
So I am to prove that cosine is continuous on R and differentiable on R. I already proved it for sine which was simple by using the identity of sin(x +- y)=sin(x)cos(y)+-cos(x)sin(y)
Now I need to prove it for cosine and also we cannot use the identity of cos(x+-y)=cos(x)cos(y)-+sin(x)(sin(y)
We do know that [itex]sin^2(x)+cos^2(x)=1[/itex] and [itex]cos(x)=1-2sin^2(x/2)[/itex]
[itex]cos(x)=sin(\pi/2 - x)[/itex], [itex]sin(x)=cos(\pi/2 - x)[/itex], [itex]cos(-x)=cos(x)[/itex], [itex]sin(-x)=-sin(x)[/itex]
sin(0)=0, cos(0)=1, [itex]|sin(x)| \le 1, |cos(x)| \le 1[/itex]
and [itex]0 < xcos(x) < sin(x) < x, 0 < x \le \pi/2[/itex]
Homework Equations
The Attempt at a Solution
So to prove that it is continuous I was thinking using [itex]cos(x)=sin(\pi/2 - x)[/itex]
so we can say that [itex]{\lim }\limits_{x \to a} cos(x)=\limits_{x \to a} sin(\pi/2 -x)=sin(\pi/2 -a)=cos(a)[/itex]
Would this be correct?
Now to prove that cos(x) is differentiable I am stumped. We do know that sin(x) is differentiable and also that [itex]{\lim }\limits_{x \to 0} sin(x)/x=1[/itex] and [itex]{\lim }\limits_{x \to 0} (1-cos(x))/x=0[/itex]
I know how to prove both using the identity cos(x+-y)=cos(x)cos(y)-+sin(x)(sin(y), but we are not allowed to do so.