Question

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!

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

nor cos for A

Answer 6

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?

Answer 7

Kind of, yes.

Answer 8

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?

Answer 9

okay, yes following so far.

Answer 10

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.

Answer 11

so.. it ends up being? (√1)?

Answer 12

Which simplifies to?

Answer 13

1..?

Answer 14

Exactly!

Answer 15

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

Answer 16

You're welcome.