Expert: Ahmed Salami - 12/5/2009

Question
what is the inverse function of y= -1/3sin(2x) and y=2cos(x/3)
and verify that it is the inverse function.

Thank you so much for your help!!!


Answer
Hi Delmi,
y = -1/3sin(2x)
-3y = sin(2x)
2x = arcsin(-3y)
x = (1/2)arcsin(-3y)
The inverse function is therefore
(1/2)arcsin(-3x)       (just change y to x)
To verify,
y = -1/3sin(2x)
x = (1/2)arcsin(-3y)
putting x in the expression for y,
y = -1/3sin[2(1/2)arcsin(-3y)]
 = -1/3sin[arcsin(-3y)]
 = -1/3(-3y)
 = y

y = 2cos(x/3)
y/2 = cos(x/3)
x/3 = arccos(y/2)
x = 3arccos(y/2)
The inverse function is therefore
3arccos(x/2)       (just change y to x)
To verify,
y = 2cos(x/3)
x = 3arccos(y/2)
putting x in the expression for y,
y = 2cos(x/3)
 = 2cos[(1/3)3arccos(y/2)]
 = 2cos[arccos(y/2)]
 = 2(y/2)
 = y

Remember, the inverse functions are;
(1/2)arcsin(-3x)  and   3arccos(x/2)

Regards