Question

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_____=______.
.

Answer 1

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 ...

Answer 2

on the interval [-90,90] sin^-1 is 1:1

For cos^-1, the domain is [0,180]

But you probably want radians instead.

Answer 3

It showed up as wrong on my sheet

Answer 4

can you see how unhelpful that response is? What did you enter, and why was it wrong?

Answer 5

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.