# What is the meaning of Trigonometry Identities being true for all values?

"Trigonometric identities hold true for all the values of $$\theta$$". I can't understand this because there are some values which Trigonometric Identities are undefined. Given the Identities:

$$\sec^2\theta-\tan^2\theta=1; |\sec\theta|\geq1$$

$$\forall\;\theta\in\mathbb{R}-\{(2n+1)\frac{\pi}{2}, n\in\mathbb{Z} \}$$

and $$\csc^2\theta-\cot^2\theta=1; |\csc\theta|\geq1$$

$$\forall\; \theta\in\mathbb{R}- \{n\pi, n\in\mathbb{Z} \}$$ The first one, for example, $$\tan\theta$$ is undefined when $$\cos\theta=0$$. Probably there exists misunderstanding about concepts by me, feel free to correct. Perhaps, this was not the best example, if there were better ones, please answer me. But I would like to know if "Trigonometric identities hold true for all the values of $$\theta$$" is always true?

• Note how the $\forall \theta \in \cdots$ part requires $\theta$ to be different from $n\pi + \pi / 2$ for any integer $n$. Or, in other words, requires $\cos \theta \neq 0$.
– chi
May 3 '20 at 8:26

$$\tan\theta=1$$ is not an identity, it is an equation. Because it is true only for certain values of $$\theta$$.

$$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ is an identity because whenever the members are defined, they are equal.

This is what is meant by "holds true for all the values of $$\theta$$".

If you prefer, there are no values of $$\theta$$ such that the two expressions differ.

• Though I can't find examples now, I've seen this called "strongly equal functions": $f(\theta),\,g(\theta)$ are defined for the same $\theta$, & are equal for all such $\theta$.
– J.G.
May 2 '20 at 18:50
• @YvesDaoust "is an identity because whenever the members are defined or undefined, they are equal". Is this true? May 2 '20 at 18:55
• @永輝123: undefined expressions cannot be compared.
– user65203
May 2 '20 at 19:02
• The alternative meaning "there are no values of θ such that the two expressions differ" is incorrect, or at least ambiguous. For $\cot\theta = \frac{1}{\tan \theta}$, the LHS is $0$ for $\theta =\frac{\pi}{2}$ (more generally, $\theta=\frac{\pi}{2}+n\pi$) while the RHS is undefined. May 3 '20 at 14:11
• @YvesDaoust: writing that one is defined, while the other is not is a comparison. I am not impressed by your "full stop". May 3 '20 at 15:27

This has to do with the principle of unique analytical continuation. In high school you prove from right triangles that for $$0 you have $$\cos^2 x+\sin^2 x=1$$. Later you learn that the functions $$x\mapsto\cos x$$ and $$x\mapsto\sin x$$ can be extended to analytic functions on all of $${\mathbb C}$$. The named principle then says that the identity $$\cos^2 x+\sin^2 x=1\qquad\left(0 enforces $$\cos^2z+\sin^2 z=1\qquad(z\in{\mathbb C})\ ,$$ and similarly for other such identities, as long as we can draw a connected region $$\Omega\subset{\mathbb C}$$ containing an arc or larger, where this identity is valid.

It is different with identities like $$\arcsin(\sin x)=x\qquad\left(-{\pi\over2}\leq x\leq {\pi\over2}\right)\ .\tag{1}$$ Here $$\arcsin$$ is not a global inverse function of $$\sin$$, but is defined ad hoc by $$(1)$$, and is the inverse of $$\sin$$ on the interval $$\bigl[-{\pi\over2},{\pi\over2}\bigr]$$. One has to make detailed studies, where in $${\mathbb C}$$ $$\arcsin$$ could be defined, and what a region $$\Omega\subset{\mathbb C}$$ could be such that $$\arcsin(\sin z)=z\qquad(z\in\Omega)\ .$$

• Thank you for your answer. So I probably got confused because of my lack of background, and it relates to the meaning of undefined as well. But there's something wrong with the examples I gave? May 2 '20 at 19:27