There is a question (no. 7) about an inverse trig function which I really don't understand how to do. I'd be really grateful if someone could explain.
C3 Trig Functions Help Watch
- Thread Starter
- 09-10-2016 15:22
- 09-10-2016 15:36
By definition, the hypothenuse has a greater length than the adjacent side, so . The ration between and will always be within , but with the unit circle, can be negative, so it will always be within , hence the domain of the function.
Given that is the inverse function of , the domain of is the range of , and the range of the first is the domain of the second.
For ii), you could draw a triangle in a unit circle where such that , and find from there.
For iii), the inverse function is a matter of flipping and and isolating .