Is $[\sin(1/2)]^{-1}$ identical to $\frac{1}{\sin(1/2)}$

Question

I'm in Grade 12 Advanced Functions and having some trouble with understanding the difference between $\sin^{-1}(1/2)$ and $(\sin(1/2))^{-1}$. I recognize that the former asks to find an angle whose sine is 1/2, and the latter is just a multiplicative inverse, and I just need this question answered to be absolutely sure in my understanding of what's going on. Thanks.

Answer 1

Yeah, the notation used in the cases of these are really really dumb, if we're being honest.

Basically, where the exponent is located is important:

• $\text{sin}^{-1}(x)$ basically denotes the "inverse sine" or "arcsine" function. It is the function you use to find the angle. For example, if you have $\text{sin}^{-1}(x)=\theta$, you see that

$$\text{sin}^{-1}(x)=\theta \;\;\; \Rightarrow \;\;\; \text{sin}(\text{sin}^{-1} (x))=\text{sin}(\theta) \;\;\; \Rightarrow \;\;\; x = \text{sin}(\theta)$$

• $[\text{sin}(\theta)]^{-1}$ denotes the multiplicative inverse of sine. Note that this isn't the same as the previous "inverse." The previous is an inverse function (more generally, $f^{-1}$ is an inverse function to $f$ if $f(f^{-1}(x)) = f^{-1}(f(x))=x$), whereas this is a multiplicative inverse. A multiplicative inverse to a number $x$ is the number $y$ such that $x\cdot y=1$, i.e. $1/x$ is the multiplicative inverse to $x$. In that sense,

$$[\text{sin}(\theta)]^{-1} = \frac{1}{sin(\theta)}$$

We note this latter result can go by a different name, the cosecant ($csc(x)$) function.

In writing out these things, to avoid any potential ambiguities, it will be helpful to you to write the first case as $arcsin(x)$ (which is an acceptable notation). The latter case could be written as a fraction or as $csc(x)$, whichever is most convenient.

Just note that, in general, the two are very, very different things that carry very different meanings ... whoever thought up the notation wasn't exactly being helpful.

Answer 2

Yes, your answer is correct. $$\sin ^{-1} (1/2) =\pi/6 $$ and $$[ \sin (1/2)]^{-1} = \frac {1}{\sin (1/2)} = \csc (1/2)$$