## anonymous 3 years ago 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!

1. anonymous

2. anonymous

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.

3. anonymous

I'm still lost.. how do you get from one to the other?

4. anonymous

Right, if A = arcsin(x) and B = arccos(x), what would the addition identity give?

5. anonymous

=arcsin(x) arccos(x) + no idea because sin is not given for b?

6. anonymous

nor cos for A

7. anonymous

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?

8. anonymous

Kind of, yes.

9. anonymous

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?

10. anonymous

okay, yes following so far.

11. anonymous

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.

12. anonymous

so.. it ends up being? (√1)?

13. anonymous

Which simplifies to?

14. anonymous

1..?

15. anonymous

Exactly!

16. anonymous

!!?! I get it! Thank you so much!

17. anonymous

You're welcome.