# Question Regarding Inverse Trig Functions

• Sep 6th 2008, 06:06 PM
RedBarchetta
Question Regarding Inverse Trig Functions
I know this isn't true:

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

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

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

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

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

Thank you.
• Sep 6th 2008, 06:12 PM
Krizalid
It depends who's workin' on it.

Some books denote $\displaystyle \sin^{-1}x=\arcsin x,$ that may cause confusion, but it's a notation addapted.
• Sep 6th 2008, 06:18 PM
RedBarchetta
Quote:

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

Some books denote $\displaystyle \sin^{-1}x=\arcsin x,$ 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.

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

Is that true? or...

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

That kinda thing. Thanks for replying Krizalid.
• Sep 6th 2008, 06:21 PM
Krizalid
Nop, they're not equal.

Assuming $\displaystyle \sin^{-1}x=\arcsin x,$ then $\displaystyle \arcsin x\ne\frac1{\text{arccsc}\,x}.$
• Sep 6th 2008, 06:29 PM
RedBarchetta
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.
• Sep 6th 2008, 07:04 PM
mr fantastic
Quote:

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

Is that true? or...

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

That kinda thing. Thanks for replying Krizalid.

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

It follows that $\displaystyle \text {cosec}^{-1} x = \sin^{-1} \frac{1}{x}$.
• Sep 6th 2008, 07:43 PM
Shyam
Quote:

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

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

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

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

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

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

Thank you.

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

In the same way,

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

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

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