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Math Help - arccos identity

  1. #1
    Super Member fardeen_gen's Avatar
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    arccos identity

    arccos [1/2.x^2 + √(1- x^2).√(1 - (x^2/4)) = arcos x/2 - arcos x, is true if:

    A)|x| ≤ 1
    B) x is a real number
    C) 0 ≤ x ≤ 1
    D)-1 ≤ x ≤ 0
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  2. #2
    Super Member fardeen_gen's Avatar
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    Can it be done this way? - I put x = any real number, x = -1, x = 1 and check validity of equation.
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  3. #3
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    Hello, fardeen_gen!

    \arccos\left(\frac{x^2}{2} + \sqrt{1- x^2}\cdot\sqrt{1 - \frac{x^2}{4}}\right) \;=\; \arccos\frac{x}{2} - \arccos x

    This is true if:

    . . (A)\;|x| \leq 1 \qquad(B)\;x\text{ is a real number} \qquad(C)\;0 \leq x \leq 1 \qquad(D)\;-1 \leq  x \leq 0
    Let: \alpha \:=\:\arccos\left(\frac{x^2}{2} + \frac{\sqrt{(1-x^2)(4-x^2)}}{2}\right) \qquad \beta \:=\:\arccos\frac{x}{2} \qquad \gamma \:=\:\arccos x

    \text{Then: }\;\cos\alpha \:=\:\frac{x^2}{2} + \frac{\sqrt{(1-x^2)(4-x^2)}}{2}\qquad\begin{array}{ccccccc}\cos\beta &=& \frac{x}{2} & \Rightarrow & \sin\beta &=&\frac{\sqrt{4-x^2}}{2} \\ cos\gamma &=& x & \Rightarrow & \sin\gamma &=&\frac{\sqrt{1-x^2}}{1} \end{array} .[1]


    We have: . \alpha \;=\;\beta - \gamma

    Take cosines: . \cos\alpha \;=\;\cos(\beta - \gamma) \;=\;\cos\beta\cos\gamma + \sin\beta\sin\gamma

    Substitute [1]: . \frac{x^2}{2} + \frac{\sqrt{(1-x^2)(4-x^2)}}{2} \;=\;\frac{x}{2}\cdot x + \frac{\sqrt{4-x^2}}{2}\cdot\frac{\sqrt{1-x^2}}{1}

    And we have: . \frac{x^2}{2} + \frac{\sqrt{(1-x^2)(4-x^2)}}{2} \;=\;\frac{x^2}{2} + \frac{\sqrt{(1-x^2)(4-x^2)}}{2}


    We have an identity . . . It is true for all values of x . . . answer (B)

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  4. #4
    Super Member fardeen_gen's Avatar
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    I have a doubt.
    In , shouldn't we keep a restriction on x so that the terms under square root don't become negative? For all real values of x, eg. 5, the terms under the sq. root become negative. But is it ok, since this happens on both sides of the equality?
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