I don't agree the sine and inverse sine functions are one to one. The sine function is cyclic.
arcsin(1/2)=30 deg and 150 and ...
I'm having a lot of trouble on this word problem. Can someone help me plz?
To define the inverse sine function, we restrict the domain of sine to the interval ______. On this interval the sine function is one-to-one, and its inverse function
sin^−1 is defined by sin^−1 x = y ⇔ sin_______=_______.
For example,
sin−1 (1/2)=_______ because sin_____=______.
(b) To define the inverse cosine function we restrict the domain of cosine to the interval_______. On this interval the cosine function is one-to-one and its inverse function
cos^−1 is defined by
cos−1 x = y ⇔ cos_______=________.
For example,
cos^−1(1/2)=_______ because cos_____=______.
.
5 answers
on the interval [-90,90] sin^-1 is 1:1
For cos^-1, the domain is [0,180]
But you probably want radians instead.
For cos^-1, the domain is [0,180]
But you probably want radians instead.
It showed up as wrong on my sheet
can you see how unhelpful that response is? What did you enter, and why was it wrong?
So basically, how would you restrict the sin/cos graph so that it becomes 1 to 1? you section off the graph via interval notations. So it would like half of a period but with its head and end cut off. (at the highest point and lowest point, so it passes the HLT - Horizontal Line Test)
... restrict the domain of cosine to the interval [-pi/2, pi/2].
you can figure the rest out.
... restrict the domain of cosine to the interval [-pi/2, pi/2].
you can figure the rest out.