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csc(acos(u))

Guest May 25, 2017
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 #1
avatar+4747 
+3

\(\csc(\arccos u ) \\~\\ =\frac{1}{\sin(\arccos u)} \\~\\ =\sqrt{ (\frac{1}{\sin(\arccos u)})^2 } \\~\\ =\sqrt{ \frac{1}{\sin^2(\arccos u)} } \\~\\\)

 

And... sin2θ = 1 - cos2θ

 

\(=\sqrt{ \frac{1}{1-\cos^2(\arccos u)} } \\~\\ =\sqrt{ \frac{1}{1-\cos(\arccos u)\cos(\arccos u)} } \\~\\ =\sqrt{ \frac{1}{1-uu} } \\~\\ =\frac{1}{\sqrt{1-u^2}} \\~\\\)

 

laugh

hectictar  May 25, 2017
 #2
avatar+90602 
+2

This is how I would do it.

 

let acos(u) = alpha  

Draw a right angled triangle and let alpha be an acute angle.

since  cos( alpha) = u/1

let  the adjacent side by u and the hypotenuse be 1

using pythagoras's theorem you get  

 \(\text{opposite side=}\sqrt{1-u^2}\)

 

\(cosec(\alpha)=\frac{hyp}{adj}=\frac{1}{\sqrt{1-u^2}}\)

 

but you do not know which quadrant alpha is in so

 

\(cosec(\alpha)=\pm\frac{1}{\sqrt{1-u^2}}\)

Melody  May 25, 2017

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