5

"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?

欲しい未来
  • 891
  • 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 03 '20 at 08:26

2 Answers2

7

$$\tan\theta=1$$ 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 02 '20 at 18:50
  • @YvesDaoust "is an identity because whenever the members are defined or undefined, they are equal". Is this true? – 欲しい未来 May 02 '20 at 18:55
  • @永輝123: undefined expressions cannot be compared. –  May 02 '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. – Taladris May 03 '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". – Taladris May 03 '20 at 15:27
2

This has to do with the principle of unique analytical continuation. In high school you prove from right triangles that for $0<x<{\pi\over2}$ 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 $$\cos^2 x+\sin^2 x=1\qquad\left(0<x<{\pi\over2}\right)$$ 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 $$\arcsin(\sin z)=z\qquad(z\in\Omega)\ .$$

Christian Blatter
  • 226,825
  • 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? – 欲しい未来 May 02 '20 at 19:27