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Math Help - Find θ for given condition?(seems to be easy but I can't seem to proceed)?

  1. #1
    Super Member fardeen_gen's Avatar
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    Find θ for given condition?(seems to be easy but I can't seem to proceed)?

    (sin θ + cos θ - 1)/(2 sin θ cos θ) = √(6) - 2

    Find θ ?
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  2. #2
    Super Member fardeen_gen's Avatar
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    I got a very messy arctan.
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  3. #3
    Like a stone-audioslave ADARSH's Avatar
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    its 15

    try to put it yaar
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  4. #4
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    Hi fardeen_gen,

    Here is my very blunt approach:

    Let's define \sin(\theta)=s~,~\cos(\theta)=c

    We have \frac{s+c-1}{2sc}=\sqrt{6}-2 Call this equation (*) Note that 2sc=\sin(2\theta)~\forall \theta \in \mathbb{Z} and thus \frac{s+c-1}{\sin(2\theta)}=\sqrt{6}-2 Now we can square both sides of the equation and do a bit of rearranging;

    \frac{(s+c-1)^2}{\sin^2(2\theta)}=(\sqrt{6}-2)^2

    \implies \frac{(s^2+c^2)+2sc-2(s+c)+1}{\sin^2(2\theta)}=6-4\sqrt{6}+4

    \implies \frac{2+\sin(2\theta)-2(s+c)}{\sin^2(2\theta)}=10-4\sqrt{6} Call this equation (**)

    Note that from equation (*) We have;

    s+c=\sin(2\theta)(\sqrt{6}-2)+1

    Now subbing this into equation (**) gives;

    \frac{2+\sin(2\theta)-2[\sin(2\theta)(\sqrt{6}-2)+1]}{\sin^2(2\theta)}=10-4\sqrt{6}

    \implies \frac{2+\sin(2\theta)-2\sin(2\theta)(\sqrt{6}-2)-2}{\sin^2(2\theta)}=10-4\sqrt{6}

    \implies \frac{1-2(\sqrt{6}-2)}{\sin(2\theta)}=10-4\sqrt{6}

    \implies \sin(2\theta)=\frac{5-2\sqrt{6}}{10-4\sqrt{6}}=\frac{5-2\sqrt{6}}{10-4\sqrt{6}}\cdot\frac{10+4\sqrt{6}}{10+4\sqrt{6}}

     <br />
\implies \sin(2\theta)=\frac{1}{2}<br />

    Maybe there is an easier 'technique'
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