Trigonometric Functions
Sine
The continuity of all other trigonometric functions can be derived as a result of the continuity of $\sin(x)$. We need two facts for this proof, which I will prove at the end of the post.
\[\lvert \cos(x) \rvert \leq 1 \qquad \forall x \in \mathbb{R}\] \[\lvert \sin( x ) \rvert = \sin(\lvert x \rvert) \leq \lvert x \rvert \qquad \forall x \in \mathbb{R}\]It also utilizes the sum-to-product identity, which I prove in my trigonometry series
\[\sin \alpha - \sin \beta = 2 \cos \left (\frac{ \alpha + \beta }{2} \right ) \sin \left (\frac{ \alpha - \beta }{2} \right )\]Fix any $\epsilon > 0$. Let $\delta = \epsilon$, then
\[\begin{align} &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad \lvert x - a \rvert < \delta \\[10pt] &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad \lvert x - a \rvert < \epsilon \\[10pt] &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad \frac{\lvert x - a \rvert}{2} < \frac{\epsilon}{2} \\[10pt] &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad \sin \left (\frac{\lvert x - a \rvert}{2} \right) < \frac{\epsilon}{2} \\[10pt] &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad \left \lvert \sin \left (\frac{ x - a }{2} \right ) \right \rvert < \frac{\epsilon}{2} \\[10pt] &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad \left \lvert \cos \left (\frac{ x + a }{2} \right ) \right \rvert \cdot \left \lvert \sin \left (\frac{ x - a }{2} \right ) \right \rvert < 1 \cdot \frac{\epsilon}{2} \\[10pt] &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad 2 \cdot \left \lvert \cos \left (\frac{ x + a }{2} \right ) \right \rvert \cdot \left \lvert \sin \left (\frac{ x - a }{2} \right ) \right \rvert< \epsilon \\[10pt] &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad \left \lvert 2 \cdot \cos \left (\frac{ x + a }{2} \right ) \cdot \sin \left (\frac{ x - a }{2} \right ) \right \rvert < \epsilon \\[10pt] &\exists \delta > 0 \quad\text{s.t.}\quad \lvert x - a \rvert < \delta \quad \implies \quad \lvert \sin (x) - \sin (a) \rvert < \epsilon \\[10pt] \end{align}\]which completes the proof.
Cosine
We use the property that $\cos(x) = \sin(\pi/2 - x)$ (which I prove in my trigonometry series) and the continuity of $\sin(x)$
\[\lim_{x \rightarrow a} \cos(x) = \lim_{x \rightarrow a} \sin(\pi/2 - x) = \sin(\pi/2 - a) = \cos(a)\]which completes the proof.
The Rest
All other trignomoetric functions can be written in terms of $\sin$ and $\cos$, so their continuity is implied by the division and reciprocal properties.
\[\lim_{x \rightarrow a} \tan(x) = \lim_{x \rightarrow a} \frac{\sin(x)}{\cos(x)} = \frac{\displaystyle \lim_{x \rightarrow a} \sin(x)}{\displaystyle \lim_{x \rightarrow a} \cos(x)} = \frac{\sin(a)}{\cos(a)} = \tan(a)\] \[\lim_{x \rightarrow a} \sec(x) = \lim_{x \rightarrow a} \frac{1}{\cos(x)} = \frac{1}{\displaystyle \lim_{x \rightarrow a} \cos(x)} = \frac{1}{\cos(a)} = \sec(a)\] \[\lim_{x \rightarrow a} \csc(x) = \lim_{x \rightarrow a} \frac{1}{\sin(x)} = \frac{1}{\displaystyle \lim_{x \rightarrow a} \sin(x)} = \frac{1}{\sin(a)} = \csc(a)\] \[\lim_{x \rightarrow a} \cot(x) = \lim_{x \rightarrow a} \frac{\cos(x)}{\sec(x)} = \frac{\displaystyle \lim_{x \rightarrow a} \cos(x)}{\displaystyle \lim_{x \rightarrow a} \sin(x)} = \frac{\cos(a)}{\sin(a)} = \cot(a)\]Inverse Trigonometric Functions
Since all of the trigonometric functions are continuous, their corresponding inverse functions must also be continuous.
Bounds on Sine and Cosine
If you look up these proofs online, you will see an argument based on the derivative of $\sin(x)$. However, I want to avoid circularity. It’s actually very subtle. If you look at my proof here, I use the fact that $\displaystyle \lim_{h \rightarrow 0} \frac{1 - \cos(h)}{h} = 0$. To prove this (given here), I use the fact that $\sin$ is continuous ($\displaystyle \lim_{x \rightarrow 0} \sin (x) = \sin(0) = 0$). Therefore, I cannot use the derivative of $\sin(x)$ to prove this bound.
Sine
I will resort to first principles and use the geometric definition of $\sin$: the height of the right triangle formed by points on the unit circle.
We will assume that the angle $x$ is in the first quadrant, i.e. that $x \in [0, \pi/2]$. The diagram shows that by definition $\sin(x)$ is the height of the right triangle. Furthermore, by definition, the length of the arc of the circle highlighted in red is equal to the angle $x$ measured in radians.
Now, we can almost assert the bound without proof. Both the line and the arc travel from the point on the unit circle to the x-axis. The line $\sin(x)$ is the shortest distance from this point to the x-axis, and therefore it must be the case that $\sin(x) \leq x$ for all $x \in [0, \pi/2]$.
Now, to get the rest of the values of $x$. By definition, we know that $\abs{\sin(x)} \leq 1$. Thus, if $\abs{x} \geq \pi/2$ it must be the case that $\abs{\sin(x)} \leq 1 < \pi/2 \leq \abs{x}$. Therefore, we have proven the result for all values of $x$.
Cosine
The bound for $\cos(x)$ comes from the definition of $\cos(x)$. It is obvious that the image of cosine cannot exceed the value of $1$.