SOLUTION: Find all solutions to sin^-1(1/2) between 0 degrees and 360 degrees. (I know that 30 degrees is a solution, but I'm not sure if there are anymore)

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Question 1174709: Find all solutions to sin^-1(1/2) between 0 degrees and 360 degrees. (I know that 30 degrees is a solution, but I'm not sure if there are anymore)
Found 2 solutions by Theo, AnlytcPhil:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
problem is:

find all solutions for angle = arcsin(1/2)

solve for angle to get:
angle = 30 degrees.

that's the angle in the first quadrant.

the sine function is positive in the first and second quadrant.
the sine function is negative in the third and fourth quadrants.

the equivalent angle in the second quadrant is 180 - 30 = 150 degrees.
the equivalent angle in the third quadrant is 180 + 30 = 210 degrees.
the equivalent angle in the fourth quadrant is 360 - 30 = 330 degrees.

all of these angles have the same value of sine, except for the sign.

to be more specific.

sin(30) = 1/2
sin(150) = 1/2
sin(210) = -1/2
sin(330) = -1/2

i believe they only want the solutions that have the arcsin as 1/2 and not the solutions that have the arcsine as -1/2.

that would restrict the answer to 30 degrees and 150 degrees.

the equation of y = sin(x) can be graphed.

the intersection of tha equation with y = .5 should show all the possible solutions between 0 and 360 degrees.

here's the graph.




Answer by AnlytcPhil(1807)   (Show Source): You can put this solution on YOUR website!
That's wrong.

Since we want inverse trig functions to be functions, they must always
have a unique value, so the rule that's chosen by mathematicians is:

Inverse sine, cosine and tangent functions are restricted so that they can only
be the smallest possible angle in absolute value, and if that results in a choice
between a positive or a negative angle, the positive angle is chosen.

So sin-1(1/2) = 30°

All other angles which have sine equaling 1/2 are larger in absolute value than
30°, so 30° is the only answer.

Edwin
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