Math

For questions 2 and 3, use a calculator to find the values of the inverse function in radians.
sin^-1(0.65)
0.71 + 2pi n and –0.71 + 2 pi n
0.71 + 2 pi n and –3.85 + 2 pi n
0.86 + 2pi n and –0.86 + 2pi n
–0.61 + 2pi n and 2.54 + 2 pi n

tan^-1(-0.09)
–0.09 + 2pi n
No such angle exists.
–1.48 + pi n
–0.09 + pi n

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1. Here is how I do these:
sin^-1 (.65)
I know from CAST that the sine is positive in quadrants I and II
so 2ndF sin .65 gave me .70758...
That is my quad I angle, so π - .70758 or 2.434 would be the quadrant II angle
my answer would be .71 + n(2π) and 2.434 + n(2π), where n is an integer
I don't see that answer but notice that to reach 2.434 they went in the negative directions, so
2.434 and -3.85 are co-terminal angles, so I guess the 2nd choice would be it.

Do the 2nd question in the same way, knowing that the tangent is negative in II and IV, also the period of the tangent function is π, so add nπ to your two answers.

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posted by Reiny

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