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Re: Inverse trigonometrical functions


On 25 Jan 2010, at 11:06, Arnold wrote:

> How by means of Mathematica to transform =
ArcSin[x*Sqrt(1-y^2)-y*Sqrt(1-x^2)] in ArcSin[x]-ArcSin[y]?
>

Note that:

TrigExpand[Sin[ArcSin[x] - ArcSin[y]]]

x*Sqrt[1 - y^2] - Sqrt[1 - x^2]*y

That means that x*Sqrt[1 - y^2] - Sqrt[1 - x^2]*y == ArcSin[x*Sqrt[1 - y^2] - Sqrt[1 - x^2]*y] provided
Abs[ArcSin[x] - ArcSin[y]]<=Pi/2 (assuming both x and y are reals between -1 and 1).

To see some examples, let

eq = ArcSin[x*Sqrt[1 - y^2] - y*Sqrt[1 - x^2]] ==
  ArcSin[x] - ArcSin[y];

then

eq /. {x -> 1, y -> -1}

False

eq /. {x -> 1, y -> 0}

True


Andrzej Kozlowski



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