Posted by anon on Thursday, March 3, 2011 at 1:35am.
Write equivalent equations in the form of inverse functions for
(can you show how you would solve)
x= y+ cos è
cos è = x-y
theta = cos^-1(x-y)
cos(y) = x^2
y = Cos^-1(x^2)
Trigonometry - MathMate, Thursday, March 3, 2011 at 8:31am
They are both correct!
Trigonometry - Reiny, Thursday, March 3, 2011 at 8:34am
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)
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
Trigonometry - MathMate, Thursday, March 3, 2011 at 1:17pm
Reiny, I interpret "inverse" being inverse trigonometric function, since he is doing a trigonometry course. However, I stand to be corrected!
Trigonometry - Reiny, Thursday, March 3, 2011 at 3:06pm
You are right.
Funny how one's mind can get stuck along one track, and other possibilities get blocked.
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