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  • calculus -

    Not much calculus here, but here goes:

    by definition
    cos(cos^-1(x)) = x
    cos^-1(cos(x)) = x

    for x in suitable ranges. Now, for cos^-1(x) the function takes on principal values between 0 and π.

    So, cos^-1(cos(15π/6)) = cos^-1(cos 5π/2) = cos^-1(0) = π/2.

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