Continuity And Differentiability

[wany]

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 $sin^2(x)+cos^2(x)=1$ and $cos(x)=1-2sin^2(x/2)$
$cos(x)=sin(\pi/2 - x)$, $sin(x)=cos(\pi/2 - x)$, $cos(-x)=cos(x)$, $sin(-x)=-sin(x)$
sin(0)=0, cos(0)=1, $|sin(x)| \le 1, |cos(x)| \le 1$
and $0 < xcos(x) < sin(x) < x, 0 < x \le \pi/2$

Homework Equations

The Attempt at a Solution

So to prove that it is continuous I was thinking using $cos(x)=sin(\pi/2 - x)$
so we can say that ${\lim }\limits_{x \to a} cos(x)=\limits_{x \to a} sin(\pi/2 -x)=sin(\pi/2 -a)=cos(a)$
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 ${\lim }\limits_{x \to 0} sin(x)/x=1$ and ${\lim }\limits_{x \to 0} (1-cos(x))/x=0$

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.

 

[micromass]

The continuity part is correct.

For the derivative, try something similar. Try deriving $$\sin(\pi/2 -x)$$ with the chain rule.

 

[wany]

Ok, so using the chain rule we know that $\pi/2 -x$ is differentiable to -1.
So by the chain rule $[cos(x)]'=[sin(\pi/2 - x)]'=[sin(\pi/2 - x)]'*(\pi/2 - x)'=[sin(\pi/2 - x)]'*(-1)$
Since sine is differentiable with sin(x)'=cos(x) then $[sin(\pi/2 - x)]'=cos(\pi/2 - x)=sin(x)$
So therefore $[sin(\pi/2 - x)]'*(-1)=sin(x)*(-1)=-sin(x)$

Is this the correct way?
And thank you so far for your help.

 

[micromass]

That seems to be correct!

 

[wany]

Cool. Thank you very much.