at 1:35am
I posted ;
Write equivalent equations in the form of inverse functions for
a.)x=y+cos theta
b.)cosy=x^2
my answers were
a.)
x= y+ cos theta
cos theta = x-y
theta = cos^-1(x-y)
b.)
cosy=x^2
cos(y) = x^2
y = Cos^-1(x^2)
your post confused me a little. Can you clarify if my answers were wrong.
See:
http://www.jiskha.com/display.cgi?id=1299215166#1299215166.1299222280
Your answers are not completely correct. Let me explain how to find the correct answers using inverse functions for each equation.
a) To find the inverse equation for the equation x = y + cos(theta), we need to isolate y. Here's the correct process:
x = y + cos(theta)
Subtract cos(theta) from both sides:
x - cos(theta) = y
Switch the sides of the equation:
y = x - cos(theta)
So the correct inverse equation is y = x - cos(theta).
b) To find the inverse equation for the equation cosy = x^2, we need to isolate y. Here's the correct process:
cosy = x^2
To isolate y, we need to take the inverse cosine (arccos) of both sides:
arccos(cosy) = arccos(x^2)
Since the range of arccos is limited, we take the principal value of arccos(x^2):
y = arccos(x^2)
So the correct inverse equation is y = arccos(x^2).
In summary, your answers for part a) and part b) were partially incorrect. The correct answers are:
a) Inverse equation: y = x - cos(theta)
b) Inverse equation: y = arccos(x^2)