Solving Trig Identity: Sin[1/2 sin^−1 (x)] = 1/2 X (sqr(1+x) –sqr(1-x))

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SUMMARY

The discussion focuses on solving the trigonometric identity: sin[1/2 sin^−1 (x)] = 1/2 * (sqrt(1+x) - sqrt(1-x)). The user, Dan, attempts to manipulate the equation by letting 2y = sin^−1(x) and subsequently using the sine double angle formula, x = sin(2y) = 2sin(y)cos(y). Despite these efforts, Dan expresses difficulty in progressing further with the solution. The conversation highlights the complexity of trigonometric identities and the need for clear notation in mathematical expressions.

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DUC_123
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Good morning, may I possibly get a clue into solving this problem : check this identity: sin[1/2 sin^−1 (x) ]=1/2 X (sqr(1+x) –sqr(1-x)). I tried to say "2y=sin^−1 (x)", Then f=sin(y), then "x=sin(2y)", x=2sin(x) X cos(x), but I can't find further. Thank you very much.
 
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DUC_123 said:
Good morning, may I possibly get a clue into solving this problem : check this identity: sin[1/2 sin^−1 (x) ]=1/2 X (sqr(1+x) –sqr(1-x)). I tried to say "2y=sin^−1 (x)", Then f=sin(y), then "x=sin(2y)", x=2sin(y) X cos(y), but I can't find further. Thank you very much.
 
Please, no x's or X's. If you must use something use *. Many people just use a space, such as in x = 2 sin(y) cos(y).

-Dan
 

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