Showing that $|\sin(a+x)-\sin(a)-\cos(a)x|\leq x^2$

I am trying to work through a small problem (finding a Fréchet Derivative), and I arrive at a function that is "obviously" less than $$x^2$$ for all $$x$$, which would be very nice to prove. I say "obviously" because by looking at the plots I visually notice that it is always smaller than $$x^2$$. But I cannot figure out how to show it rigorously.

Here is the function, with $$a$$ some constant real number.

$$f_a(x) = |\sin(a+x)-\sin(a)-\cos(a)x|$$

I have tried to rewrite the expression with some trigonometric identities, like so: $$f_a(x) = |2\sin(x/2)\cos(a + x/2) - \cos(a)x|$$ But this gets me nowhere. The fact that the whole expression is inside absolute values means I cannot really do much with it, unless I use the triangle inequality to get $$f_a(x) \leq |2\sin(x/2)\cos(a + x/2)| + |\cos(a)x|$$but this does not seem strict enough to show the upper limit of $$x^2$$.

How would I go about showing something like this?

• what are you allow to use? Do you allow to use $\sin' = \cos$ using the usual definition on $\mathbb R$? – user251257 Jun 12 '20 at 15:37
• I suppse I am allowed to use anything that I understand myself. Are you talking about using the mean value theorem here somehow? – BodyDouble Jun 12 '20 at 15:51
• yes, or a variant of taylor theorem. – user251257 Jun 12 '20 at 15:52

For $$a\in\mathbb{R}$$, let $$g_a(x) = \sin(a+x)-\sin(a)-\cos(a)x \qquad\qquad\qquad\qquad\qquad\qquad\;\;\;$$ Fix $$a,x\in\mathbb{R}$$.
Claim:$$\;$$If $$x\ne 0\;$$then $$|g_a(x)| < x^2$$.
First suppose $$x > 0$$. \begin{align*} \text{Then}\;\,g_a(x) &=\bigl(\sin(a+x)-\sin(a)\bigr)-\cos(a)x\\[4pt] &=\cos(a+t)x-\cos(a)x\;\text{for some t\in (0,x)}&&\text{[by the MVT]}\\[4pt] &=x\bigl(\cos(a+t)-\cos(a)\bigr)\\[4pt] &=x\bigl(t(-\sin(a+s)\bigr)\;\text{for some s\in (0,t)}&&\text{[by the MVT]}\\[6pt] \text{hence}\;\,|g_a(x)|&=|x||t||-\sin(a+s)|\le |x||t| < x^2 \end{align*} Next suppose $$x < 0$$. \begin{align*} \text{Then}\;\,g_a(x) &=\bigl(\sin(a+x)-\sin(a)\bigr)-\cos(a)x\\[4pt] &=\cos(a+t)x-\cos(a)x\;\text{for some t\in (x,0)}&&\text{[by the MVT]}\\[4pt] &=x\bigl(\cos(a+t)-\cos(a)\bigr)\\[4pt] &=x\bigl(t(-\sin(a+s)\bigr)\;\text{for some s\in (t,0)}&&\text{[by the MVT]}\\[6pt] \text{hence}\;\,|g_a(x)|&=|x||t||-\sin(a+s)|\le |x||t| < x^2 \end{align*} This completes the proof.