cos^-1 (x^2/2 + √(1 - x^2)√(1 - (x^2/4))) = cos^-1(x/2) - cos^-1 x is valid for which value of x ?

Question

\(cos^{-1}\Bigg(\cfrac{x^2}2+\sqrt{1-(\cfrac{x^2}4)}\Bigg)\) = \(cos^{-1}(\cfrac{x}2)-cos^{-1}x\) is valid for which value of x ?

With the help of:

Answer 1

The domain of cos⁻¹ x is [−1, 1]

We know that cos^-1x - cos^-1y = cos^-1(xy + \(\sqrt{1-x^2}\sqrt{1-y^2}\)).

If x ∈ [0, 1], then

  \(cos^{-1}(\cfrac{x}2)-cos^{-1}x\) = \(cos^{-1}\Bigg(\cfrac{x^2}2+\sqrt{1-(\cfrac{x^2}4)}\Bigg)\)for x ∈ [0, 1].

 Therefore, \(cos^{-1}\Bigg(\cfrac{x^2}2+\sqrt{1-(\cfrac{x^2}4)}\Bigg)\) = \(cos^{-1}(\cfrac{x}2)-cos^{-1}x\) is not valid x ∈ [0, 1].

If x ∈  [−1, 0], then replacing x by – x, we get

Therefore, \(cos^{-1}\Bigg(\cfrac{x^2}2+\sqrt{1-(\cfrac{x^2}4)}\Bigg)\) = \(cos^{-1}(\cfrac{x}2)-cos^{-1}x\) is not valid x ∈ [-1, 0].

Hence, \(cos^{-1}\Bigg(\cfrac{x^2}2+\sqrt{1-(\cfrac{x^2}4)}\Bigg)\) = \(cos^{-1}(\cfrac{x}2)-cos^{-1}x\) is valid for x ∈ [0, 1].