A few Inverse questions (1 Viewer)

[imoO]

Hey guys,

I'm having trouble working these few questions out:

sin^-1(2sin(-pi/6)) =

tan^-1(3tan(7pi/6)) =

sin^-1 = tan^-1 (x/sqrt (1-x^2))

Sorry if my notation is no good. They are all questions from the Cambridge 3-U yr 12 textbook.

Secondly,

I have a few questions from past papers.

if f(x)=3x-x^3 how do i fink the gradient of the inverse function at x=0? Obviously I know how to find the inverse function, but I'm having trouble making y the subject so I can differentiate. My mediocre maths isn't helping.....

Lastly, How do i sketch f(x) = (e^x) / (3+e^x). I hate complicated functions like these.....

Thanks for any help in advanced,

 

[Drongoski]

sin^-1(2sin(-pi/6)) = sin^-1(2 x (-0.5)) = sin^-1(-1) = -pi/2

tan^-1(3 tan(7pi/6)) = tan^-1(3 x tan(pi/6)) = tan^-1(3 x (1/rt(3))) = tan^-1(rt(3)) = pi/3


Don't get Question no 3.


RE: y = f(x) = 3x - x^3, if f has an inverse function, then gradient of inverse is dx/dy

and dx/dy = 1/(dy/dx) = 1/(3 - 3x^2) = 1/3 @ x=0 (if I'm not mistaken)

 

[exiting]

Question 3, perhaps draw the triangle up.

 

[study-freak]

Q3.
sin^(-1) ?= tan^(-1) (x/sqrt (1-x^2))

Construct a right-angled triangle with an gle theta and adjacent side sqrt(1-x^2), opposite side x, hypotenuse 1.

Then tan^(-1) (x/sqrt (1-x^2))=theta=sin^(-1) ?
sin theta=x=?

The ? mark represents the variable or constant you didn't write.