Sin(2sin-1(x/3)) stuck at last step

[jwxie]

Homework Statement

Simplify this expression.

sin(2* arcsin(x/3))

Homework Equations

The Attempt at a Solution

Let t = arcsin(x/3), we get sin(2t) which is double angle
sin(2T) = 2 sinT cosT

I did cosT first

T = arcsin x/3, and we know inverse of arcsin is sin T = x/3
this gives us the hypotenuse 3, and opposite side = x (3sin delta = x)
using the identity of cos^2 + sin^2 = 1, and solve for cos, I get cosT = sqrt(9-x^2)

This is corrected.

Now, solve for sinT.

sin(arcsin(x/3)), is the same as just x/3, by definition I think.
Well, the answer the book gave was 2x/9 * cosT
I get the cosT part right, but now the x/9

I don't understand how he produced the number 9.

Thank you. I have an exam so please pardon me posting so many questions these two days.

 

[happyg1]

You said that you used sin^2+cos^2=1 to solve for the cos T. It should be \cos T= \sqrt{1-sin^2 T}
\cos T=\sqrt {1-\frac{x^2}{3^2}}. Find a common denominator under the radical and a 3 pops out in the denominator.

 

[jwxie]

I still don't get it.
Right, the identity is that, so in this case, we have sin^2 + cos^2 = 3^2
so cost is right (what i wrote)
i still can't understand how we get 1/9 outside for sinT

 

[happyg1]

The identity doesn't have 3^2, it has 1 \sin^2 T + \cos ^2 T=1 , not 9.

You are mixing up the the identity with the triangle.

So far you have \frac{2x}{3}\sqrt{1-\frac{x^2}{9}
Then you find the common denominator under the radical:

\frac{2x}{3}\sqrt{\frac{9-x^2}{9}}=\frac{2x}{3}\frac{\sqrt{9-x^2}}{3}

 

[jwxie]

The identity doesn't have 3^2, it has 1 \sin^2 T + \cos ^2 T=1 , not 9.

You are mixing up the the identity with the triangle.

So far you have \frac{2x}{3}\sqrt{1-\frac{x^2}{9}
Then you find the common denominator under the radical:

\frac{2x}{3}\sqrt{\frac{9-x^2}{9}}=\frac{2x}{3}\frac{\sqrt{9-x^2}}{3}

Brilliant. Thank you for pointing this out.
My math professor only repeats what the book prints, and my background in Math is a bit weak.
Thanks for all the help!