Inverse of trigo. functions

arc sin2x = arc sin(sqrt3)x + arc sinx

sin(arc sin2x) = sin(arc sin(sqrt3)x + arc sinx)

2x = sin(sin(sqrt3)x)cos(arc sinx) + cos(sin(sqrt3)x)sin(arc sinx)

2x = (sqrt3)xsqrt(1 - x^2) + xsqrt(1 - 3x^2)

x(sqrt(3 - 3x^2) + sqrt(1 - 3x^2) - 2) = 0

x = 0 or sqrt(3 - 3x^2) + sqrt(1 - 3x^2) - 2 = 0

sqrt(3 - 3x^2) = 2 - sqrt(1 - 3x^2)

3 - 3x^2 = 4 - 4sqrt(1 - 3x^2) + (1 - 3x^2)

sqrt(1 - 3x^2) = 1/2

1 - 3x^2 = 1/4

x^2 = 1/4

x = +- 1/2

So, x = 0 or +- 1/2


2009-10-18 22:07:30 補充：
Just bear in mind

sin(a + b) = sina cosb + cosa sinb

這裏可以幫到你
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