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!
Hello, Higg!
Do you know what a geometric mean is?
$\displaystyle \text{4. Find one geometric mean between: }\:2\sin^2\!\theta,\:\_\_\!\_\!\_\, ,\:\,\cos^2\!\theta\,\text{ in terms of }\sin\theta\text{ only.}$
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: .$\displaystyle x \;=\;\sqrt{2}\sin\theta\cos\theta \quad\Rightarrow\quad x \;=\;\sqrt{2}\sin\theta\sqrt{1-\sin^2\!\theta}$
Hello again, Higg!
I have two issues with this problem.
This is what you posted.
$\displaystyle \text{4. Find one geometric mean between: }\:{\color{red}2}\sin^2\!\theta, \;\_\_\_\_\,,\:\cos^2\!\theta \text{ in terms of }{\color{blue}\sin\theta}\text{ only.}$
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 sin2θ.