Show that if $\lim_{x\to a} f(x) = L$, then $\lim_{x \to a} cos(f(x))=cos(L)$.
Hint: looking up some trigonometric identities may help you
I have been using delta-epilson to prove this statement, but the hint that has been given is throwing me off. Why would using trigonometric identities help with this question?
$$\cos(A)-\cos(B)=-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$$ Apply this with $A=f(x)$ and $B=L$. Then $$\left|\cos(f(x))-\cos(L)\right|=2\left|\sin\left(\frac{f(x)+L}{2}\right)\sin\left(\frac{f(x)-L}{2}\right)\right|.$$ Now, recall that $|\sin(t)|\le|t|$, so, $$\left|\cos(f(x))-\cos(L)\right|\le\frac{1}{2}|f(x)+L||f(x)-L|.$$ And from here, it is easy to conclude a $\delta-\varepsilon$ type proof.
For the sake of completeness, $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$ $$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$$ by substracting $$\cos(a+b)-\cos(a-b)=-2\sin(a)\sin(b).$$ Now, calling $A=a+b$ and $B=a-b$, gives $a=\frac{A+B}{2}$ and $b=\frac{A-B}{2}$.