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katlynsbirds' question at Yahoo! Answers regarding inverse trigonometric identity

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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: katlynsbirds' question at Yahoo! Answers regarding inverse trignometric identity

Hello katlynsbirds,

We are given to prove:

\(\displaystyle \cot^{-1}(x)=\sin^{-1}\left(\frac{1}{\sqrt{1+x^2}} \right)\)

Let's let \(\displaystyle \theta=\cot^{-1}(x)\,\therefore\,x=\cot(\theta)\), and now please refer to this diagram:

katlynsbirds.jpg

We see that \(\displaystyle \cot(\theta)=\frac{x}{1}=x\) and we can also see that:

\(\displaystyle \sin(\theta)=\frac{1}{\sqrt{1+x^2}}\,\therefore\, \theta=\sin^{-1}\left(\frac{1}{\sqrt{1+x^2}} \right)\)

and so we may conclude:

\(\displaystyle \theta=\cot^{-1}(x)=\sin^{-1}\left(\frac{1}{\sqrt{1+x^2}} \right)\)

Shown as desired.

To katlynsbirds and any other guests viewing this topic I invite and encourage you to post other trigonometry problems here in our Trigonometry forum.

Best Regards,

Mark.