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Math Help - Question Regarding Inverse Trig Functions

  1. #1
    Member RedBarchetta's Avatar
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    Question Regarding Inverse Trig Functions

    I know this isn't true:

    <br />
\sin ^{ - 1} x \ne \frac{1}<br />
{{\sin x}}<br />

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

    <br />
\sin x = \frac{1}<br />
{{\csc x}}<br />

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

    <br />
\sin ^{ - 1} x = \frac{1}<br />
{{\csc ^{ - 1} x}}<br />

    Thank you.
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  2. #2
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    Krizalid's Avatar
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    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.
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  3. #3
    Member RedBarchetta's Avatar
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    Quote Originally Posted by Krizalid View Post
    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.

    <br />
\sin ^{ - 1} x = \frac{1}<br />
{{\csc ^{ - 1} x}}<br />


    Is that true? or...

    <br />
\tan ^{ - 1} x = \frac{1}<br />
{{\cot ^{ - 1} x}}<br />

    That kinda thing. Thanks for replying Krizalid.
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  4. #4
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    Nop, they're not equal.

    Assuming \sin^{-1}x=\arcsin x, then \arcsin x\ne\frac1{\text{arccsc}\,x}.
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  5. #5
    Member RedBarchetta's Avatar
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    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.
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  6. #6
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    Quote Originally Posted by RedBarchetta View Post
    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...

    <br />
\tan ^{ - 1} x = \frac{1}<br />
{{\cot ^{ - 1} x}}<br />

    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}.
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  7. #7
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    Quote Originally Posted by RedBarchetta View Post
    I know this isn't true:

    <br />
\sin ^{ - 1} x \ne \frac{1}<br />
{{\sin x}}<br />

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

    <br />
\sin x = \frac{1}<br />
{{\csc x}}<br />

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

    <br />
\sin ^{ - 1} x = \frac{1}<br />
{{\csc ^{ - 1} x}}<br />

    Thank you.
    Let <br />
\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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