Inverse Trig Function question (1 Viewer)

[redorange]

This question comes from the 1991 3u HSC paper Q 7)b)

The solutions that came with this past paper book didnt really make sense for part (iii)

So can anyone else use another method to solve (iii) ?

Thanks

http://img135.imageshack.us/img135/1410/trig3ch.gif

 

[redorange]

Ah thats the same as the solutions in the book.

Whats this compound angle thing?

 

[pLuvia]

According to (ii) the maximum angle is formed when x=2
Maximum when x=2
θ=tan^-14/x-tan^-11/x
θ=tan^-12-tan^-11/2
tanθ=tan[tan^-12-tan^-11/2]
Let a=tan^-12 and b=tan^-11/2
tanθ=tan(a-b)
RHS=tan(a-b)
=(tana-tanb)/(1+tana*tanb)
=(2-1/2)/(1+2*1/2)
=3/4
tanθ=3/4
θ=tan^-13/4
QED

 

[redorange]

Oh I remember now,

Thanks peoples

 

[redorange]

Yeah, thanks

I was lookin at my other notes and i thought it was the 'angle between 2 curves' formula, so it confused me.... they look the same!