Using trig identities we can see that $\sin^2 x + \cos^2 x = \tanh^2 x + \text{sech}^2 x = 1$ , and so the parametric graph $(\cos t, \sin t)$ is similar to $(\text{sech} t, \tanh t)$. The first graph produces a semicircle when $-\frac \pi2 \le t \le \frac \pi2$ and the second graph when $-\infty \lt t \lt \infty$. My question is if it is possible to convert between the graphs; e.g. if there is a function $f$ where $(\text{sech} t, \tanh t) = (\cos f(t), \sin f(t))$.
[Math] Conversion between trig functions and hyperbolic trig functions
circlesfunctionsgraphing-functionsparametrictrigonometry
Related Solutions
Note that $$(\cos(t)+i\sin(t))^n=(\cos(nt)+i\sin(nt)),~~n\in\mathbb Z$$ and $(a+b)^3=a^3+3a^2b+3ab^2+b^3,~~~(a-b)^3=a^3-3a^2b+3ab^2-b^3$.
I created the images that "inspired" the illustration of the Angle-Sum and Difference identities on Wikipedia, so I'm all in favor of learning those relations. However, this is how I remember your particular class of identities.
First, in the Unit Circle, I imagine the "sine-cosine-$1$" triangle for a generic first-quadrant angle $\theta$, taking $\theta$ small enough that the triangle is considerably wider than it is tall, creating an obvious "long leg" and "short leg". Because trig values are positive in the first quadrant, I can say that the "long leg" of the triangle has length $\cos\theta$, and that the "short leg" has length $\sin\theta$.
Then, I imagine what happens when I rotate that triangle through multiples of $90^\circ$, getting a nice windmill:
The "(co-)reference triangle" for each of the compound angles is a rotation of the original triangle, and we can read off sine and cosine values by paying attention to the positions of the short legs and long legs (and assigning signs, as appropriate). For instance, the point labeled "$\theta+90^\circ$" is at distance "$\cos\theta$" above the $x$-axis, and at distance "$\sin\theta$" to the left of the $y$-axis; therefore, $$\sin\left(\theta+\phantom{1}90^\circ\right) = \phantom{-}\color{red}{\cos\theta} \qquad\qquad \cos\left(\theta+\phantom{1}90^\circ\right) = -\color{blue}{\sin\theta}$$
Likewise, $$\sin\left(\theta+180^\circ\right) = -\color{blue}{\sin\theta} \qquad\qquad \cos\left(\theta+180^\circ\right) = -\color{red}{\cos\theta}$$ $$\sin\left(\theta-\phantom{1}90^\circ\right) = -\color{red}{\cos\theta} \qquad\qquad \cos\left(\theta-\phantom{1}90^\circ\right) = \phantom{-}\color{blue}{\sin\theta}$$
Similarly, there's a windmill for compound angles involving $-\theta$:
And we have $$\sin\left(\phantom{180^\circ}-\theta\right) = -\color{blue}{\sin\theta} \qquad\qquad \cos\left(\phantom{180^\circ}-\theta\right) = \phantom{-}\color{red}{\cos\theta}$$ $$\sin\left(\phantom{1}90^\circ-\theta\right) = \phantom{-}\color{red}{\cos\theta} \qquad\qquad \cos\left(\phantom{1}90^\circ-\theta\right) = \phantom{-}\color{blue}{\sin\theta}$$ $$\sin\left(180^\circ-\theta\right) = \phantom{-}\color{blue}{\sin\theta} \qquad\qquad \cos\left(180^\circ-\theta\right) = -\color{red}{\cos\theta}$$ $$\sin\left(-90^\circ-\theta\right) = -\color{red}{\cos\theta} \qquad\qquad \cos\left(-90^\circ-\theta\right) = -\color{blue}{\sin\theta}$$
It's worthwhile to point out that the identity $$\cos\theta = \sin\left(90^\circ - \theta\right)$$ is, for some (such as myself), definitional:
The co-sine of the angle is the sine of the co-angle.
where "co-angle" means "complementary angle", just as "co-sine" literally means "complementary sine".
Also, the identities
$$\cos(-\theta) = \cos\theta \qquad \sin(-\theta) = -\sin\theta$$
have significance in establishing that cosine is an "even function" (it acts on negative arguments ---killing the sign--- the way an even exponent would) and sine is an "odd function" (it acts on negative arguments ---preserving the sign--- the way an odd exponent would). These are handy properties, which say interesting things about how the graphs are drawn. So, you might want to reserve a special area of your brain-space for them.
Best Answer
You want the Gudermannian function: it is defined by $$ \operatorname{gd}{x} = \int_0^x \operatorname{sech}{y} \, dy $$ In particular, this can be expressed, by changing variables, as $$ \operatorname{gd}{x} = (\operatorname{sgn}{x}) \arccos{(\operatorname{sech}{x})} = \arcsin{(\tanh{x})}, $$ so $$ \cos{(\operatorname{gd}{x})} = \operatorname{sech}{x},\\ \sin{(\operatorname{gd}{x})} = \tanh{x}, $$ and then everything follows as you wish, since $\operatorname{gd}:(-\infty,\infty) \to (-1,1)$ bijectively.