Angle-sum identities for $\csc$, $\sec$, $\cot$; $\rm{sech}$, $\sinh$, $\coth$; $\arcsin$, $\rm{arctanh}$, $\rm{arccoth}$?

Question

I'm sure many of you are aware of the following identities: $$\begin{align} \sin(A \pm B) &= \sin A\cos B \pm \sin B\cos A \\[4pt] \cos(A \pm B) &= \cos A\cos B \mp \sin A\sin B \\[4pt] \tan(A \pm B) &= \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B} \end{align}$$

I was wondering if there were any identities for other trigonometric functions such as:

1. The reciprocal trigonometric functions such as: $$\csc(A \pm B),\;\sec(A \pm B),\;\text{and}\;\cot(A \pm B)$$
2. The hyperbolic trigonometric functions such as: $$\operatorname{sech}(A \pm B),\;\sinh(A \pm B),\;\text{and}\; \coth(A \pm B)$$
3. The inverse trigonometric functions such as: $$\arcsin(A \pm B),\;\operatorname{arctanh}(A \pm B),\;\text{and}\;\operatorname{arccoth}(A \pm B)$$

Answer 1

For part $1$ and $2$, there's very lucid ways to derive these identities using Euler's identity, which I consider fundamental to trigonometry more generally. Euler's identity is: $$e^{ix} = \cos x + i\sin x$$

For example, let's consider $e^{x+y} = e^x e^y = (\cos x + i\sin x)(\cos y + i\sin y)$, where $Re(e^{x+y}) = \cos(x+y)$ and $Im(e^{x+y}) = \sin(x+y)$. Once you know these two values, you can find all other (spherical) trigonometric functions of $x+y$ in terms of the identities for $\sin$ and $\cos$, since all other functions are defined in terms of $\sin$ and $\cos$. For $\tan$, it would be $\sin/\cos$, etc.

Let's take the example $$\sec(x+y) = 1/\cos(x+y) = \frac{1}{\cos x\cos y - \sin x \sin y}$$ To simplify this expression, dividing the expression by $\cos x\cos y$ in the numerator and denominator can appear to help, so that

$$\sec(x+y) = \frac{\sec x \sec y}{1-\tan x\tan y}$$ which is the commonly accepted identity for $\sec (x+y)$. In the same spirit

$$\csc(x+y) = \frac{1}{\sin x\cos y + \cos x\sin y}$$ which can be simplified by multiplying the expression on the numerator and denominator by $\csc x\csc y$. Hence $$\csc(x+y) = \frac{\csc x\csc y}{\cot x + \cot y}$$

The most interesting however is the case of $\cot(x+y)$, which is the easiest to write just in terms of $\cot x$ and $\cot y$. From the identity for $\tan(x+y)$ it follows that $$\cot(x+y) = \frac{1-\tan x\tan y}{\tan x + \tan y}$$ Multiplying the expression by $\cot x\cot y$ on both the numerator and denominator gives $$\cot(x+y) = \frac{\cot x\cot y - 1}{\cot x + \cot y}$$

This also connects very well to hyperbolic trigonometry. From the definition of $2\sinh(x) = e^x - e^{-x}$ it follows that $\frac{\sin ix}{i}$ is just $\sinh x$. Similarly, since $2\cosh(x) = e^x + e^{-x}$, it follows that $\cos(ix)$ is just $\cosh x$. From the identity of $\sin(x+y)$ it follows that

$$\sin(ix+iy) = \sin(ix)\cos(iy) + \cos(ix)\sin(iy)$$ which upon division by $i$ yields the identity

$$\sinh(x+y) = \sinh x\cosh x + \cosh y\sinh y$$ From the identity for $\cos(x+y)$ it follows that $$\cos(ix+iy) = \cos(ix)\cos(iy) - \sin(ix)\sin(iy)$$ Dividing and multiplying the second term by $i^2 = -1$ in the right hand side gives $$\cosh(x+y) = \cosh x\cosh y + \sinh x \sinh y$$

Part $3$ is more difficult. Let's say we wish to work out $\arcsin(x+y)$. Then suppose $x+y = \sin a$ so that $\arcsin(x+y) = a$. We immediately find that there's no nice way to write $a$ in terms of $\arcsin(x)$ and $\arcsin(y)$ just from the information that $\sin(a) = x+y$. However, there are identities for $\arcsin x + \arcsin y$, etc. As an example, for $\arcsin x + \arcsin y$, suppose that $\sin a = x$ and $\sin b = y$, so that you're looking for $a+b$. Then note that $\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$ and hence $a+b = \arcsin x + \arcsin y = \arcsin(x\sqrt{1-y^2} + y\sqrt{1-x^2})$, but depending on the values of $x,y$ you need to take care of which quadrant $\arcsin x + \arcsin y$ lies in, since the principal value branch for $\arcsin(k)$ has the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Answer 2

Multiple questions here. Not really in policy. In response I answer only question 1.

There are such identities, but they are seldom used because we may conveniently render the cotangent, secant and cosecant as respectively the reciprocals of the tangent, cosine and sine.

We can use these reciprocal relationships to construct the missing identities, but the results are unwieldy for the secant and cosecant functions. The cotangent case gives a more workable outcome and is the one explored here.

Begin with the tangent udentity and takes its reciprocal to obtain

$\cot(A\pm B)=\dfrac{1\mp\tan A\tan B}{\tan A\pm\tan B}$

and simply multiply the numerator and denominator by $\cot A\cot B$ to exchange the tangets for cotangents:

$\cot(A\pm B)=\dfrac{\cot A\cot B\mp1}{\cot B\pm\cot A}.$

Note the reversed order of the arguments in the denominator. We have $\tan A\cot A\cot B=\cot B$ and vice versa.