anonymous
  • anonymous
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!
Mathematics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Right, if A = arcsin(x) and B = arccos(x), what would the addition identity give?
anonymous
  • anonymous
=arcsin(x) arccos(x) + no idea because sin is not given for b?
anonymous
  • anonymous
nor cos for A
anonymous
  • 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?
anonymous
  • anonymous
Kind of, yes.
anonymous
  • 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?
anonymous
  • anonymous
okay, yes following so far.
anonymous
  • 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.
anonymous
  • anonymous
so.. it ends up being? (√1)?
anonymous
  • anonymous
Which simplifies to?
anonymous
  • anonymous
1..?
anonymous
  • anonymous
Exactly!
anonymous
  • anonymous
!!?! I get it! Thank you so much!
anonymous
  • anonymous
You're welcome.

Looking for something else?

Not the answer you are looking for? Search for more explanations.