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# trig

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csc(acos(u))

Guest May 25, 2017
#1
+7323
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$$\csc(\arccos u ) \\~\\ =\frac{1}{\sin(\arccos u)} \\~\\ =\sqrt{ (\frac{1}{\sin(\arccos u)})^2 } \\~\\ =\sqrt{ \frac{1}{\sin^2(\arccos u)} } \\~\\$$

And... sin2θ = 1 - cos2θ

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

hectictar  May 25, 2017
#2
+93604
+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