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Inverse trigonometry prove this

  1. Jan 10, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    If [itex]cos^{-1}\frac{x}{a}+cos^{-1}\frac{y}{b} = \alpha[/itex] then show that [itex]\frac{x^2}{a^2}-\frac{2xy}{ab}cos \alpha + \frac{y^2}{b^2} = sin^2 \alpha[/itex]


    2. Relevant equations

    3. The attempt at a solution
    I assume inverse functions to be θ and β respectively. So in the second equation for LHS I can convert it like this

    [itex] cos^2\theta + cos^2\beta - 2cos \alpha cos\theta cos\beta [/itex]

    Similarly for RHS
    [itex] sin^2 \alpha = sin^2(\theta + \beta)[/itex]
     
  2. jcsd
  3. Jan 12, 2013 #2
    Taking the inverse cosine of a cosine gives the argument to the inverse cosine. Then you would use the sum of angles trigonometric formula.
     
  4. Jan 12, 2013 #3

    SammyS

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    So you need to show that [itex]\ \ \cos^2(\theta) + \cos^2(\beta) - 2\cos(\theta + \beta) \cos(\theta)\cos(\beta)= \sin^2(\theta + \beta)\ .[/itex]


    [itex]\cos(\theta+\beta)=\cos(\theta)\cos( \beta)-\sin(\theta)\sin(\beta) [/itex]


    [itex]\displaystyle \sin^2(\theta+\beta)=\left(\sin(\theta)\cos(\beta)+\cos(\theta)\sin(\beta) \right)^2\\
    \quad\quad\quad=\sin^2(\theta)\cos^2(\beta)+2\sin(\theta)\cos(\beta)\cos(\theta)\sin(\beta)+\cos^2( \theta)\sin^2(\beta)
    [/itex]

    So you can get some cancellation.


    BTW: Didn't you post this very same question previously?
     
  5. Jan 13, 2013 #4

    utkarshakash

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    No. Not exactly like this.
     
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