Sketching inverse trig graph question (1 Viewer)

[+GriM ReApeR+]

sketch y=cos^-1(cosx)

sketch y=cosx(cos^-1(x))

Also, evaluate tan^-1(-(square root)3)...is the answer no sloution because the solutions are not in the range?

 

[grimreaper]

the first one is just y=x

Did you mean y=cos(cos^-1(x)) (I think you put an extra x in there accidently). Thats just y = x as well

Oh and youll have to think about domain\range in both of those.

And yes, there is a solution to the last one (there is a solution to any value of x)

 

[kimmeh]

Originally posted by +GriM ReApeR+
Also, evaluate tan^-1(-(square root)3)...is the answer no sloution because the solutions are not in the range?

when you type it into the calculator, its -60

Originally posted by +GriM ReApeR+
sketch y=cos^-1(cosx)

cos^-1 and cos will "cancel out" to give you a graph y = x. however, this graphs isnt just y=x. substitute values in and from the top of my head, you should get like a sawtooth looking graph.

Originally posted by +GriM ReApeR+
sketch y=cosx(cos^-1(x))

you might want to try multiplication of ordinates