Here's the question you clicked on:
Atkinsoha
PreCalc & Trig Question! (Picture below) please try to be as specific as possible.. as I have not been able to go to school for over a week and have missed all of these lessons!
Try using the identity \[\sin(A+B)=\sin(A)\cos(B)+\sin(B)\cos(A)\]Then \[\sin^2(x)+\cos^2(x)=1\] See how far you get with that.
I'm still lost.. how do you get from one to the other?
Right, if A = arcsin(x) and B = arccos(x), what would the addition identity give?
=arcsin(x) arccos(x) + no idea because sin is not given for b?
Huh? Let: \[A=\sin^{-1}(x) \textrm{ and } B=\cos^{-1}(x)\]Then using the sum identity that I wrote: \[\sin(\sin^{-1}(x) + \cos^{-1}(x)) = \sin(\sin^{-1}(x))\cos(\cos^{-1}(x))+\sin(\cos^{-1}(x))\cos(\sin^{-1}(x))\]Follow up to here?
Ok.. From here, we know that sin(arcsin(x)) = x and cos(arccos(x)) = x, so: \[ = x^2 + \sin(\cos^{-1}(x))\cos(\sin^{-1}(x))\] Now you can write \[\sin(x)=\sqrt{1-\cos^2(x)}\]and\[\cos(x)=\sqrt{1-\sin^2(x)}\]Using the second identity that I wrote. Again, do you follow?
okay, yes following so far.
Ok, so we now get \[=x^2 + \sqrt{1-\cos^2(\cos^{-1}(x))}\sqrt{1-\sin^2(\sin^{-1}(x))}\]\[=x^2 + \sqrt{1-x^2}\sqrt{1-x^2}\] Hopefully you can do the rest.. If there's anything you don't get, say so.
so.. it ends up being? (√1)?
!!?! I get it! Thank you so much!