Question Regarding Inverse Trig Functions

**RedBarchetta** · September 6th 2008, 06:06 PM

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.

**Krizalid** · September 6th 2008, 06:12 PM

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.

**RedBarchetta** · September 6th 2008, 06:18 PM

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.

**Krizalid** · September 6th 2008, 06:21 PM

Nop, they're not equal.

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

**RedBarchetta** · September 6th 2008, 06:29 PM

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.

**mr fantastic** · September 6th 2008, 07:04 PM

Originally Posted by RedBarchetta

Yes. I get the arcsin notation, my question is about the last equality I wrote.
$ 

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

**Shyam** · September 6th 2008, 07:43 PM

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,

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

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

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

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