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sin^-1 [(radical 3)/2] (in the first quadrant)
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basically find sin rad3/2 in the first quadrant in the unit circle
cos are the x values sin is the y values in it
so i would be looking at 60 degrees in the first quadrant?
yup
well what do i do with it, is that all? isn't sin prime does it alter the answer?
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60 degrees or pie/3 is your answer, there the same thing, not sure what you mean with the second part of your question
in my problem it is typed as: sin^-1 is that the same as the original sin?
no its the inverse
when given a inverse problem i always say where in the unit circle is this value located
you just happen to be limited to quadrant 1 for your problem
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oh ok thank you
np
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