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AussieDave
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[SOLVED] Question involving trigonometric identities and inverse functions
http://img141.imageshack.us/img141/2651/quiz1question5zi8.jpg
I've tried to combine the following known equations to come up with a solution:
[tex]\frac{d}{dx}[/tex](sin[tex]^{-1}[/tex]x) = [tex]\frac{1}{\sqrt{1-x^{2}}}[/tex]
sin x = y
sin [tex]^{-1}[/tex] y = x
[tex]\frac{d}{dx}[/tex](sin x) = cos x
cos[tex]^{2}[/tex]x + sin[tex]^{2}[/tex]x = 1
I've been writing down a whole bunch of different equations on a sheet of paper to try and come up with something to connect the two equations. I feel like I'm kind of shooting in the dark though as I'm not sure where to begin and how to use this knowledge of the derivatives (if that's needed) and the relationship between cos and sin. I've tried fiddling around with the pythagorean identity but I end up with things like:
x = sin[tex]^{-1}[/tex]x[tex]\sqrt{1-cos^{2}x}[/tex]
and I'm not sure where to go from there.
Your help will be much appreciated.
Kind regards,
David
Homework Statement
http://img141.imageshack.us/img141/2651/quiz1question5zi8.jpg
Homework Equations
I've tried to combine the following known equations to come up with a solution:
[tex]\frac{d}{dx}[/tex](sin[tex]^{-1}[/tex]x) = [tex]\frac{1}{\sqrt{1-x^{2}}}[/tex]
sin x = y
sin [tex]^{-1}[/tex] y = x
[tex]\frac{d}{dx}[/tex](sin x) = cos x
cos[tex]^{2}[/tex]x + sin[tex]^{2}[/tex]x = 1
The Attempt at a Solution
I've been writing down a whole bunch of different equations on a sheet of paper to try and come up with something to connect the two equations. I feel like I'm kind of shooting in the dark though as I'm not sure where to begin and how to use this knowledge of the derivatives (if that's needed) and the relationship between cos and sin. I've tried fiddling around with the pythagorean identity but I end up with things like:
x = sin[tex]^{-1}[/tex]x[tex]\sqrt{1-cos^{2}x}[/tex]
and I'm not sure where to go from there.
Your help will be much appreciated.
Kind regards,
David
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