Trig of inverse trig or?

Geometry Level 3

$\large \color{#0C6AC7}{\sin(} \color{teal}{\sin^{-1}(x)} \color{#0C6AC7} {)} \qquad \left|\right |\qquad \color{teal}{\sin^{-1}(} \color{#0C6AC7}{\sin(x)} \color{teal} {)}$

If $x\, \epsilon \, \mathbb R$ , then what is the difference between the two functions above?

They have different ranges One of them is an even function They are same They have different domains

1 solution

Md Omur Faruque
Jul 14, 2015

Although,, both of the function is equal to $x$ , $\color{teal} {\sin^{-1}(x) }$ is defined only between $x\epsilon [-1,1]$ . So, $\color{#0C6AC7}{\sin(} \color{teal}{\sin^{-1}(x)} \color{#0C6AC7} {)}$ has a domain of $[-1,1]$ .

Whereas, being a periodic function $\color{#0C6AC7} {\sin(x) }$ is defined for any real number of x, where $\sin x \epsilon [-1,1]$ . So, $\color{teal}{\sin^{-1}(} \color{#0C6AC7}{\sin(x)} \color{teal} {)}$ is defined for any $x\epsilon R$ .

Henceforth, the only correct answer is, $\color{#69047E} {\boxed {\text{They have different domains}}}$

Moderator note:

I've edited your problem slightly since you are only taking real numbers into consideration.

Bonus question : What would the answer be if $x$ can be a complex number?

if they have different domains , they also have different ranges .... correct me if i'm wrong

Anand O R - 5 years, 11 months ago

No, both of the functions have a range of $\boldsymbol {[-1,1]}$

MD Omur Faruque - 5 years, 11 months ago

okay . now i got it

Anand O R - 5 years, 11 months ago

Trig of inverse trig or?

1 solution

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