Trigonometry
posted by anon .
Write equivalent equations in the form of inverse functions for
a.)x=y+cos è
b.)cosy=x^2
(can you show how you would solve)
a.)
x= y+ cos è
cos è = xy
theta = cos^1(xy)
b.)
cosy=x^2
cos(y) = x^2
y = Cos^1(x^2)

They are both correct!

in a) is è a variable or a constant.
If it is a variable then we don't find an "inverse"
(the inverse is found by interchanging the x and y variables in a 2 variable relation)
in b)
original : cosy = x^2
inverse: cosx = y^2
solving this for y:
y = ±√cosx,
solving this for x:
x = cos^1 (y^2) , where 1 < y < 1 
Reiny, I interpret "inverse" being inverse trigonometric function, since he is doing a trigonometry course. However, I stand to be corrected!

You are right.
Funny how one's mind can get stuck along one track, and other possibilities get blocked.