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 June 13th, 2010, 07:07 PM #1 Newbie   Joined: Jun 2010 Posts: 1 Thanks: 0 How are these the first three positive values of theta? (Inv For the inverse function, arccos(4/5)=4(theta-40*), I have to find the three first positive values of theta. Simplified, I got the equations: 49.217+(?n/2) and 30.7825+(?n/2). Using these equations, I got the three positive values: 30.7823, 32.3533, and 49.2174. However, the book says the first three positive values are 30.7823* (got that one),49.2174*(got that one too), and 120.783* (didn't get this one).
June 14th, 2010, 09:33 AM   #2
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Re: How are these the first three positive values of theta? (Inv

Hello, Noraowolf!

[color=red]skipjack is right . . . I blew it![/color]

Quote:
 $\arccos\left(\frac{4}{5}\right) \:=\:4(\theta-40^o)$ Find the first three positive values of $\theta.$ The answers are:[color=beige] .[/color]$30.7823^o,\;49.2174^o,\;120.783^o$

$\text{We have: }\;4\theta\,-\,160^o \;=\;\arccos(0.8)$

[color=beige]. . . . . . . . [/color]$4\theta \,-\,160^o \;=\;\pm36.8699^o\,+\,360^on$

[color=beige]. . . . . . . . . . . . . [/color]$4\theta \;=\;160^o \, \pm\,36.8699^o\,+\,360^on$

[color=beige]. . . . . . . . . . . . . . [/color]$\theta \;=\;40^o \,\pm\, 9.2175^o\,+\,90^on$

[color=beige]. . . . . . . . . . . . . . [/color]$\theta \;=\;\begin{Bmatrix}49.2175^o \\ 30.7825^o\end{Bmatrix}\,+\,90^on$

$\text{The first three positive values are: }\;\begin{array}{cc} 30.7825^o \\ 49.2175^o \\ 120.7825^o \end{array}$

 June 14th, 2010, 11:10 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,927 Thanks: 2205 That is incorrect. $4(\theta \,-\,40^\circ)\,=\,\pm36.8699^\circ\,+\,360^\circ n$ $\theta\,=\,(\pm36.8699^\circ)/4\.+\,90^\circ n\,+\,40^\circ\,=\,\pm9.2175^\circ\,+\,90^\circ n\,+\,40^\circ.$ Hence the first few positive values are 30.7825°, 49.2175°, 120.7825° and 139.2175°.

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