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

  1. #1
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    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!
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  2. #2
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    Re: Find a geometric mean (sorry guys, not sure how to describe.)

    Hello, Higg!

    Do you know what a geometric mean is?


    \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 x such that: . 2\sin^2\!\theta:x \;=\;x: \cos^2\!\theta

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

    Therefore: . x \;=\;\sqrt{2}\sin\theta\cos\theta \quad\Rightarrow\quad x \;=\;\sqrt{2}\sin\theta\sqrt{1-\sin^2\!\theta}
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    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, Find a geometric mean (sorry guys, not sure how to describe.)-question4.jpg

    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
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    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.

    \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.
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  5. #5
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    Re: Find a geometric mean (sorry guys, not sure how to describe.)

    Quote Originally Posted by Higg View Post
    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 2\sin^2(x)\cos^2(x)=\tfrac{1}{2}\sin^2(2x).
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