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I know what the graphs look like (my book has them in the answers) but I don't know how to get there. That's the problem.

Hey guys, I thought I might post the questions here rather than create a new thread on the same topic. I too am having difficulties sketching some inverse graphs: 1. y = 2sin-1(1-x2) 2. y = 4cos-1sqrt(1-x2) 3. y = cos-1(cosx) Need some help these are really driving me crazy!

Can anyone assist me with this simple differentiation?: d/dx sin(tan-1x) I get (cos(tan-1x))/(1+x2) The answer is different. Cheers

Ok thanks that was a weird q for me. But i kinda get it.

Ty for verifying that. I'm really sorry to bother you guys but i have another question: Evaluate: tan-1x + tan-11/x I don't understand this q, when you simplify it, you get a zero in the denominator.

Sigh, another question: sin-1(2sqrt2)/3 - tan-12 I get sin-1((2sqrt2)-2)/3sqrt5 The answer is tan-1(18-10sqrt2)/31 So can there be two answers to this question? i.e. you can use sin(alpha - beta) or you could use tan(alpha - beta) right?

Edit: nevermind i get it now. :)

Oh thanks. I did not know that this applied.

hmm i am still little confused. I drew a triangle with sides 1 and 2 and hypotenuse sqrt5. One of the sides has to be negative. How do you work out which side? I'm prob not getting something (i am a bit slow unfortunately ;)

Another one for the smart kids: cos(tan-1(-2)) i drew up the triangle, but how do you know which side of the triangle to make negative?

I might hijack this thread for a bit. Can anyone help me find the exact value of: cos-1(cos 7pi/6) I'm a little confused. I solved the inside of the brackets (cos 210) which equals -cos30. But this is not in the domain of 0<(or = to)x<(or = to)pi. How can i fix this?

Sometimes it's not your idea that may be poor. Writers are their own worst enemies sometimes.

Cheers man. lyounamu<SCRIPT type=text/javascript> vbmenu_register("postmenu_3441969", true); </SCRIPT> <!-- Added by James for Benefactors: 11.30pm, 8 Feb 2005 --> <!-- Added by James for BigBloggers: 18 April 2006 --><!-- -->Completing PhD in counselling & helping...

The answer in the book is: g-1(x) = 1-sqrt(1-x2)/x (i assume they just divided everything by 2) I wonder why they only took the negative root and left out the other consideration?

Oh so the quadratic formula was needed? Well i'll be damned. Cheers mark

Maybe I just haven't had enough sleep or something but I can't seem to make "y" the subject in the following (this is an inverse trig question): x = 2y/(1+y2) Help is appreciated, thanks!

Oh I'm here alright. I just lurk around. ;)

Jeez thanks man. You are my saviour. :D

Another problem. Same question different part: 1. Given that x = asin(nt+α), find v as a function of time (<--this part i can do). Find a, n and α if a>0, n>0, 0≤α<2pi and: c) the period is 2pi seconds and initially x=1 and v=-1. Here's what i've done: T=2pi=2pi/n n=1 when t=0...

Correct me if I'm wrong but isn't most contemporary stuff subversion some way or another?