Evaluate the following expressions. Your answer must be an angle in radians and in the interval [(-pi/2),(pi/2)] .
(a) sin^(-1)(0) =
(b) sin^(-1)((-sqrt3)/2) =
(c) sin^(-1)(-1/2) =
arcsin(0) = 0
sin < 0 in QIV, so -pi/3
ditto, so -pi/6
(a) Well, we know that the clown's favorite sin is zero. So, if we apply the clown's perspective to sin^(-1)(0), we find that the angle must be in the interval [(-pi/2), (pi/2)]. Therefore, sin^(-1)(0) is equal to 0. In other words, the clown takes zero sins. Hurray for moral purity!
(b) Ah, the square root of three divided by two. This is a tricky one! The clown has to put on his thinking hat...or rather, his thinking wig! After careful clown calculations, it turns out that sin^(-1)((-sqrt3)/2) is equal to -pi/3. So, in clown terms, sin^(-1)((-sqrt3)/2) is like eating negative pi/3 slices of pie, which is quite a dessert choice!
(c) Sin^(-1)(-1/2) needs some clown logic applied to it. The clown reckons that if you have negative half the sinniest sin, you'll end up with a negative angle. And after some juggling around, it turns out that sin^(-1)(-1/2) equals -pi/6. So, the clown would say that sin^(-1)(-1/2) is like finding yourself stuck in a negative pi/6 spin cycle.
(a) To evaluate sin^(-1)(0), we need to find an angle whose sine is 0. Since the sine of 0 is 0, we can conclude that sin^(-1)(0) is equal to 0 radians.
(b) To evaluate sin^(-1)((-sqrt3)/2), we need to find an angle whose sine is (-sqrt3)/2. From the unit circle, we know that the sine of (pi/3) is (√3)/2. Taking the inverse sine, we get sin^(-1)((√3)/2) = (pi/3). However, we are looking for an angle whose sine is the negative of (√3)/2. In the unit circle, this occurs in the fourth quadrant. Therefore, sin^(-1)((-√3)/2) = -(pi/3) radians.
(c) To evaluate sin^(-1)(-1/2), we need to find an angle whose sine is -1/2. From the unit circle, we know that the sine of -(pi/6) is -1/2. Therefore, sin^(-1)(-1/2) = -(pi/6) radians.
To evaluate these expressions, we need to find the inverse sine (also called arcsine) of the given input values.
(a) To find `sin^(-1)(0)`, we need to find the angle whose sine is 0.
The sine function is 0 at two angles: 0 radians, and pi radians (180 degrees).
However, we are asked to give the answer in the interval [(-pi/2),(pi/2)], which means we need to choose the angle within that range.
In this case, the angle in that range is 0 radians, since it lies between -pi/2 and pi/2.
Therefore, `sin^(-1)(0) = 0` radians.
(b) To find `sin^(-1)((-sqrt3)/2)`, we need to find the angle whose sine is (-sqrt3)/2.
The sine function is equal to (-sqrt3)/2 at -pi/3 radians (-60 degrees).
However, we are asked to give the answer in the interval [(-pi/2),(pi/2)].
In this case, the angle in that range is -pi/3 radians, since it lies between -pi/2 and pi/2.
Therefore, `sin^(-1)((-sqrt3)/2) = -pi/3` radians.
(c) To find `sin^(-1)(-1/2)`, we need to find the angle whose sine is -1/2.
The sine function is equal to -1/2 at -pi/6 radians (-30 degrees).
Again, we are asked to give the answer in the interval [(-pi/2),(pi/2)].
In this case, the angle in that range is -pi/6 radians.
Therefore, `sin^(-1)(-1/2) = -pi/6` radians.