Hint: use the identity ...
Hello, DylanRenke!
Let: .
The problem becomes: . .(a)
From [1]: /
We have: .
We have a right triangle with:
Pythagorus tells us that: .
Hence: . .(b)
From [2]: .
We have: .
We have a right triangle with:
Pythagorus tells us that: .
Hence: . .(c)
Substitute (b) and (c) into (a): .
This is wrong. "arccos(x)/arcsin(x)" is NOT "arccot(x)". Just because a function has some property doesn't mean its inverse has that property!
Notice that and so=x
Thank you
However, cosine and sine are not "one to one" functions so they do NOT, strictly speaking, have inverses. In order to be able to talk about "arccos(x)" and "arcsin(x)" we hage to restrict cosine to 0 to and sine to to [tex]\pi/2[/itex]. With those restrictions, sine and cosine are both postive and so both and are equal to [tex]\sqrt{1- x^2}[tex].