Asked 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 = x-y
theta = cos^-1(x-y)
b.)
cosy=x^2
cos(y) = x^2
y = Cos^-1(x^2)
a.)x=y+cos è
b.)cosy=x^2
(can you show how you would solve)
a.)
x= y+ cos è
cos è = x-y
theta = cos^-1(x-y)
b.)
cosy=x^2
cos(y) = x^2
y = Cos^-1(x^2)
Answers
Answered by
MathMate
They are both correct!
Answered by
Reiny
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
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
Answered by
MathMate
Reiny, I interpret "inverse" being inverse trigonometric function, since he is doing a trigonometry course. However, I stand to be corrected!
Answered by
Reiny
You are right.
Funny how one's mind can get stuck along one track, and other possibilities get blocked.
Funny how one's mind can get stuck along one track, and other possibilities get blocked.
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