Question

Trig help please!
(Problem in comments)

Answer 1

In the expression \[\sqrt{9-u ^{2}}\] let u = 3sinx. What is the resulting expression?
I know the answer is \[3\left| cosx \right|\] but I have no idea how to get there.

Answer 2

Forget about the square root sign for now, remember that 3^2=9

Let u = 9sin^2(x)

BUT

sin^2(x)+cos^2(x)=1

Mutliply everything by 9

9sin^2(x)+9cos^2(x)=9
9-9sin^2(x)=9cos^2(x)

Subbing back in:

sqrt(9cos^2(x))
3cos(x).

Does this make sense?
If you need a further explanation I can do that.

Answer 3

If you let u=3sin(x)
Then u^2=9sin^2(x)

Answer 4

Okay I think I got it, thanks!