u can use a triangle...2 is the hypotenuse and -root3 is a side so the other side must be 1,ok i get that... just not wher to put it... can u do tan[arcsin(-root3/2)]
i no i can do with calc... just working confusing
thanks heaps... that makes it very clear... i cant believe its that obviousy = sin x is a function as it passes the vertical line test (for every x there is only one y). However it fails the horizontal line test, which implies that for every y-value there are multiple x-values.
This means the inverse relation (where you swap x and y) has multiple y-values for every x-value (notice this is swapping x and y from bold statement).
For it to be an inverse FUNCTION (a function has one y-value for every x-value), you have to restrict the domain for y = sin x, say -π/2 ≤ x ≤ π/2. This is equivalent to restricting the range of y = sin-1x to -π/2 ≤ y ≤ π/2. Therefore we can have an inverse FUNCTION in the given restrictions because the 1-1 property exists. If we did not restrict it, there would be no 1-1 correspondance between x and y.
yer ok ill figure it out thanksu can use a triangle...2 is the hypotenuse and -root3 is a side so the other side must be 1,
so tan(-root3/1)=-60
I think you would do the same thing you would do for normal functions, ie usehow would i show if y=sinx and y=arcsinx is an odd or even function.
what you need to know is ASTC, and that positive angles are read anti-clockwise on the unit circle, and negative angles are measured clock-wise on the unit circle. So you could just figure it out through that^ i forgot about tht.. and ive forgotten trig.... 4 which ratios will ratio (-x) = -ratio(x)
u can but theres easier way to do it whihc i did today lolwhat you need to know is ASTC, and that positive angles are read anti-clockwise on the unit circle, and negative angles are measured clock-wise on the unit circle. So you could just figure it out through that