Thread: algebraically find arcsine?

hi, this might be a simple problem but I can not figure it out and my textbook does not describe how to algebraically find arcsine or any arc function.

its says arcsin(-1/2) = sin(-pi/6) the angle whose sine is -1/2.

but it does not describe how to algebraically go from arcsin to sin(-pi/6)

can someone show how to do this? i dont want to use the calculator!@#

thanks

Originally Posted by skoker

hi, this might be a simple problem but I can not figure it out and my textbook does not describe how to algebraically find arcsine or any arc function.

its says arcsin(-1/2) = sin(-pi/6) the angle whose sine is -1/2.

but it does not describe how to algebraically go from arcsin to sin(-pi/6)

can someone show how to do this? i dont want to use the calculator!@#

thanks

arcsin(-1/2) = -pi/6, not arcsin(-1/2) = sin(-pi/6). You are expected to know sin and cos of 'special angles' in the first quadrant and then, using thses angles as 'reference angles' and using the symmetry of the unit circle, to be able to calculate the sin and cos of 'special angles' in the other quadrants.

In this case, you are expected to know that sin(pi/6) = 1/2.

thanks!

is this the general way to find these exact angles? just remember all the angles. seems a bit shabby.

Some students don't understand the meaning of inverse trigonometrical functions, i.e: Arcsin(x)...

Like to the function f(x)=x^2 has an inverse function which is g(x)=sqrt(x)... so to the function S(x)=sin(x) has it inverse function,which is a(x)=arcsin(x).

(Of course not to all function has inverse function, but I leave it...)

Originally Posted by skoker

thanks!

is this the general way to find these exact angles? just remember all the angles. seems a bit shabby.

There is a recent thread in this subforum that answers this question. Personally, I suggest you memorise them.

ok, I will look for the thread thanks guys.