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2006 paper II Q10(b) - HSN forum

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2006 paper II Q10(b)


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#1 Ron90

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Posted 16 May 2008 - 03:50 PM

Im stuck on this question.

Ive looked at a solution to it and it goes like this:

25cos(x-1.29) = 1

then to

x = cos^-1\frac{1}{25} + 1.29

im confused how this is done. Any help?

#2 Marcus

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Posted 16 May 2008 - 03:59 PM

\cos{x} and \cos^{-1}{x} are inverse functions of each other, ie

y = \cos{x} can be changed to x = \cos^{-1}{y} and vice versa
you started with
25\cos(x-1.29)=1

divide through by 25
\cos(x-1.29)=\frac{1}{25}

take inverse cosines of both sides ie, use the rule above
x-1.29=\cos^{-1}(\frac{1}{25})

then move the 1.29 to the other side
x=\cos^{-1}(\frac{1}{25})+1.29

hope this helps biggrin.gif
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#3 Ron90

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Posted 16 May 2008 - 04:02 PM

QUOTE(Marcus @ May 16 2008, 04:59 PM) View Post
\cos{x} and \cos^{-1}{x} are inverse functions of each other, ie

y = \cos{x} can be changed to x = \cos^{-1}{y} and vice versa
you started with
25\cos(x-1.29)=1

divide through by 25
\cos(x-1.29)=\frac{1}{25}

take inverse cosines of both sides ie, use the rule above
x-1.29=\cos^{-1}(\frac{1}{25})

then move the 1.29 to the other side
x=\cos^{-1}(\frac{1}{25})+1.29

hope this helps biggrin.gif



Brilliant. Get it now thanks smile.gif






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