Thread: Question Regarding Inverse Trig Functions

1. Question Regarding Inverse Trig Functions

I know this isn't true:

$
\sin ^{ - 1} x \ne \frac{1}
{{\sin x}}
$

Yet, does the same rules apply to inverse functions regarding the reciprocals? For example:

$
\sin x = \frac{1}
{{\csc x}}
$

Still my main question remains; Is this equality true? Can I apply the same known trigonometric reciprocals with the inverse trig functions?

$
\sin ^{ - 1} x = \frac{1}
{{\csc ^{ - 1} x}}
$

Thank you.

2. It depends who's workin' on it.

Some books denote $\sin^{-1}x=\arcsin x,$ that may cause confusion, but it's a notation addapted.

3. Originally Posted by Krizalid
It depends who's workin' on it.

Some books denote $\sin^{-1}x=\arcsin x,$ that may cause confusion, but it's a notation addapted.
Yes. I get the arcsin notation, my question is about the last equality I wrote.

$
\sin ^{ - 1} x = \frac{1}
{{\csc ^{ - 1} x}}
$

Is that true? or...

$
\tan ^{ - 1} x = \frac{1}
{{\cot ^{ - 1} x}}
$

That kinda thing. Thanks for replying Krizalid.

4. Nop, they're not equal.

Assuming $\sin^{-1}x=\arcsin x,$ then $\arcsin x\ne\frac1{\text{arccsc}\,x}.$

5. Ah ha! That's what I thought. I searched through the book but it didn't answer this question. I'm assuming that when you read the chapter you are supposed see that the domains of these functions are limited; and because of this, you cannot apply the same rules to them as you do with regular trig functions.

Thanks again.

6. Originally Posted by RedBarchetta
Yes. I get the arcsin notation, my question is about the last equality I wrote.
$
\sin ^{ - 1} x = \frac{1}

{{\csc ^{ - 1} x}}
" alt="
\sin ^{ - 1} x = \frac{1}

{{\csc ^{ - 1} x}}
" />

Is that true? or...

$
\tan ^{ - 1} x = \frac{1}
{{\cot ^{ - 1} x}}
$

That kinda thing. Thanks for replying Krizalid.
$\text {cosec}^{-1} x = \theta \Rightarrow \text {cosec} \, \theta = x \Rightarrow \frac{1}{\sin \theta} = x \Rightarrow \sin \theta = \frac{1}{x} \Rightarrow \theta = \sin^{-1} \frac{1}{x}$.

It follows that $\text {cosec}^{-1} x = \sin^{-1} \frac{1}{x}$.

7. Originally Posted by RedBarchetta
I know this isn't true:

$
\sin ^{ - 1} x \ne \frac{1}
{{\sin x}}
$

Yet, does the same rules apply to inverse functions regarding the reciprocals? For example:

$
\sin x = \frac{1}
{{\csc x}}
$

Still my main question remains; Is this equality true? Can I apply the same known trigonometric reciprocals with the inverse trig functions?

$
\sin ^{ - 1} x = \frac{1}
{{\csc ^{ - 1} x}}
$

Thank you.
Let $
\text {cosec}^{-1} x = \theta \Rightarrow \text {cosec} \, \theta = x \Rightarrow \frac{1}{\sin \theta} = x \Rightarrow \sin \theta = \frac{1}{x} \Rightarrow \theta = \sin^{-1} \frac{1}{x}$

In the same way,

$\boxed {sec^{-1}x = cos^{-1} \left( \frac{1}{x}\right)}\; \Rightarrow \; \boxed {cos^{-1}x = sec^{-1} \left( \frac{1}{x}\right)}$

$\boxed {cot^{-1}x = tan^{-1} \left( \frac{1}{x}\right)}\; \Rightarrow \; \boxed {tan^{-1}x = cot^{-1} \left( \frac{1}{x}\right)}$

$\boxed {cosec^{-1}x = sin^{-1} \left( \frac{1}{x}\right)}\; \Rightarrow \; \boxed {sin^{-1}x = cosec^{-1} \left( \frac{1}{x}\right)}$