Find a geometric mean (sorry guys, not sure how to describe.)

Hey, sorry I can't describe further in the title. I have no clue where to go with this haha.

4. Find one geometric mean between 2sin^2θ , ___ , cos^2θ in terms of sinθ only.

Thanks!

Re: Find a geometric mean (sorry guys, not sure how to describe.)

Hello, Higg!

Do you know what a geometric mean is?

Quote:

$\displaystyle \text{4. Find one geometric mean between: }\:2\sin^2\!\theta,\:\_\_\!\_\!\_\, ,\:\,\cos^2\!\theta\,\text{ in terms of }\sin\theta\text{ only.}$

We want an expression $\displaystyle x$ such that: .$\displaystyle 2\sin^2\!\theta:x \;=\;x: \cos^2\!\theta$

We have: .$\displaystyle \frac{2\sin^2\!\theta}{x} \;=\;\frac{x}{\cos^2\!\theta} \quad\Rightarrow\quad x^2 \:=\:2\sin^2\!\theta\cos^2\!\theta$

Therefore: .$\displaystyle x \;=\;\sqrt{2}\sin\theta\cos\theta \quad\Rightarrow\quad x \;=\;\sqrt{2}\sin\theta\sqrt{1-\sin^2\!\theta}$

1 Attachment(s)

Re: Find a geometric mean (sorry guys, not sure how to describe.)

Thanks!

I understand how you tackled it, except the last part, from x =... to x=....

Also, Attachment 26558

That is what I have in my Solutions key. It's a little different, so could you maybe explain that one to me as well?

Big thanks : D

Re: Find a geometric mean (sorry guys, not sure how to describe.)

Hello again, Higg!

I have two issues with this problem.

This is what you posted.

Quote:

$\displaystyle \text{4. Find one geometric mean between: }\:{\color{red}2}\sin^2\!\theta, \;\_\_\_\_\,,\:\cos^2\!\theta \text{ in terms of }{\color{blue}\sin\theta}\text{ only.}$

First, what happened to the 2 ?

In the solution, they solved a different problem.

Second, it says, "in terms of sinθ only".

But their answer contains sin**2**θ.

Re: Find a geometric mean (sorry guys, not sure how to describe.)

Quote:

Originally Posted by

**Higg** That is what I have in my Solutions key. It's a little different, so could you maybe explain that one to me as well?

It could be noted that $\displaystyle 2\sin^2(x)\cos^2(x)=\tfrac{1}{2}\sin^2(2x)$.