How to symplify arcsin(sin(9pi/7)) ?

Thank You

1 Answer

  • 4 weeks ago

    arcsin returns numbers in the range [-pi/2, pi/2]

    Since 9pi/7 is out of that range, we need to find the argument within that range that will return the same sin value.

    If you know the unit circle then you obviously use pi - 9pi/7 = -2pi/7

    sin(9pi/7) = sin(-2pi/7)

    Since -2pi/7 is in the range of the arcsin function,

    arcsin(sin(-2pi/7)) = -2pi/7

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