Difficult Trig Problem

[Bigglesworth95]

Hello,

I'm having a difficult time with this trig problem as I review for an upcoming exam. Any help with it would be fantastic and I would be really very appreciative . Please see a screen cap of the problem attached, but I will also right it out below in anyone prefers to view it that way.

Instructions: Draw a right triangle to simplify the given expression. (Hint: Use Sin 2*theta = 2(sin theta * cos theta)

sin(2cos^-1x) = ?

Thanks in advance to anyone who wants to help out with this! It means a lot!

 

[MarkFL]

Using the given double-angle identity for sine, we may write:

[MATH]\sin(2\arccos(x))=2\sin(\arccos(x))\cos(\arccos(x))=2x\sin(\arccos(x))[/MATH]
Now, draw a right triangle with an acute angle \(\theta\), and the leg adjacent to this angle label \(x\) and the hypotenuse 1. Thus we have:

[MATH]\theta=\arccos(x)[/MATH]
What then must \(\sin(\theta)\) be?

 

[Bigglesworth95]

Then sin must be 1-x². Thank you so much for the help! I totally forgot about that double angle identity.

 

[MarkFL]

We would have:

[MATH]\sin(\theta)=\sqrt{1-x^2}[/MATH]
Hence:

[MATH]\sin(2\arccos(x))=2x\sqrt{1-x^2}[/MATH]