Solving for x and y in sin-1x=y and siny=x, and tan-1x=y and tany=x

[xpack]

http://img194.imageshack.us/img194/1383/68225760.png [/URL]
sin^-1x=y & siny=x
tan^-1x=y & tany=x
The answer is E but I don't know how my teacher got to that. Can someone explain this to me please?

Homework Statement

 

[n!kofeyn]

Do you understand the inverse trig functions? The inverse trig functions find the angle of the triangle that satisfies the given relation. That is sin^-1(t) is the angle theta such that sin(theta)=t. If you view t as t/1, then can you set up a triangle that has this property? Can you then find the length of the remaining edges so that you can find tan(theta), i.e. tan(sin^-1(t))?

 

[xpack]

I'm still very confused

 

[n!kofeyn]

Okay. Set θ=sin^-1(t). This means that sin(θ)=t (i.e. sin(θ)=t/1). This means that in a right triangle with one angle that is θ, the side opposite θ has length t and the hypotenuse has length 1. You can find the side adjacent to θ by using the Pythagorean theorem. Then all you have to do is find tan(θ) since tan(θ)=tan(sin^-1(t)).

 

[xpack]

Okay, thank you so much you cleared that right up!

 

[n!kofeyn]

No problem! That is the general technique when doing these type of trig problems that contain inverse trig functions.