Is it valid for hyperbolic trigonometric identities to be proven using their relationship to the circular trigonometric functions?

We know that $$\cos(\theta)=\cosh(i\theta)$$ and $$\sin(\theta)=-i\sinh(i\theta)$$. I have figured out and have been told on various other posts and websites that these relationships can be used to prove identities involving hyperbolic trigonometric functions simply by converting them to their circular forms, applying the corresponding circular identity, and then reconverting. For example, if I wanted to prove that $$\cosh(2\theta)=\cosh^2(\theta)+\sinh^2(\theta)$$, then: \begin{align}\cosh(2\theta)&=\cos(2i\theta)\\&=\cos^2(i\theta)-\sin^2(i\theta)\\&=\cos^2(i\theta)+(-i\sin(i\theta))^2\\&=\cosh^2(\theta)+\sinh^2(\theta)\end{align}

I understand that this can be used to prove the other identities involving the hyperbolic functions. My question is whether it is valid to do this (and the reason why) because the circular trigonometric identities are only proven for real angles as far as I'm aware, while we are using imaginary angles in these proofs. (For example, I have assumed that the double-angle formula for cosine holds for imaginary angles in my above proof.) A rigorous proof or a brief reason would be fine depending on how complicated the answer is.

The reason this is important for me is that I want to use this technique for my exams (as it is much easier and more intuitive than using exponential form), but I'm unsure whether I'll be marked down for proofs such as these if they are not mathematically valid or rigorous.

I hope that was clear enough. This is my first post to this stack exchange.

• Yes, it is valid. Unless you are restricted so that you cannot use complex numbers. Commented Jan 4, 2020 at 13:36

The functions $$\cos z,\,\sin z,\,\cosh z,\,\sinh z$$ converge everywhere in $$\Bbb C$$ to their Taylor series, so results proven algebraically for real arguments extend naturally. This is applicable, for example, when you're proving compound-argument formulae (and consequences thereof such as prosthaphaeresis formulae), because they're derivable from two matrix groups,$$\left(\begin{array}{cc} \cos x & -\sin x\\ \sin x & \cos x \end{array}\right)=\exp\left(\begin{array}{cc} 0 & -x\\ x & 0 \end{array}\right)$$and$$\left(\begin{array}{cc} \cosh x & \sinh x\\ \sinh x & \cosh x \end{array}\right)=\exp\left(\begin{array}{cc} 0 & x\\ x & 0 \end{array}\right).$$With something like $$\lim_{x\to0}\frac{\sin x}{x}=1$$, which is proven with the squeeze theorem by considering the geometry of acute-angle sectors of circles, I can imagine readers demanding a more careful transition to $$\lim_{x\to0}\frac{\sinh x}{x}=1$$. You can probably see how you can argue that these limits have to be the same by comparing the small-$$x$$ behaviour of these matrices. Alternatively, you would need to argue the meromorphic function $$\frac{\sin z}{z}$$ will have the same $$z\to0$$ limit from all directions in $$\Bbb C$$.

However, these minutiae can be avoided by proving hyperbolic results afresh from$$\cosh x:=\frac{\exp x+\exp -x}{2},\,\sinh x:=\frac{\exp x-\exp -x}{2}.$$

• Thanks for the explanation. Commented Jan 5, 2020 at 11:43

As you observe, you've only see the proof of $$\sin^2 u + \cos^2 u = 1 \tag{eq. 1}$$ for real numbers $$u$$; you're applying it (in your proof) for non-real numbers $$u$$.

As it happens, eq. 1 does hold for all complex values $$u$$, and once you've established that (as @J.G. describes, using complex taylor series, for example), you can certainly use it to prove the things you're hoping to prove.

But if you have not yet established that, a grader for your exam might reasonably claim that you've reduced the cosh-sinh problem to a different unsolved problem (namely, proving eq. 1 holds for all complex numbers) and were therefore not done with your work.

I'd therefore discuss this with the professor before attempting this approach.

• Okay will do. Thanks. Commented Jan 5, 2020 at 11:43