Question

What did I do wrong with the inverse sin
(see attached)

Answer 1

Ok, so lets look at them one by one.

Answer 2

The inverse sin would be be zero, or pi, or 2pi, or 3pi, or...etc.

Answer 3

arcsin (-1/2) is telling you that there is an angle who's sine is -1/2. The sine function has starts at 0 when the angle is 0, and then goes to 1 when the angles is pi/2. So you need the angle that is between 0 and pi/2, which is pi/4. So:\[\sin^{-1} (-\frac{ 1 }{ 2 })=\frac{ \pi }{ 4 }\]

Answer 4

Sorry, \[\sin^{-1} (-\frac{ 1 }{ 2 })=-\frac{ \pi }{ 4 }\]

Answer 5

the last one is pi/4

Answer 6

The second one. I already said that the sine function starts at 0 when the angle is 0, so\[\sin^{-1} 0=0\]

Answer 7

Now, the third one is the hardest. First we note that we are looking for an angle which give a negative sine, so it should be between -pi/2 and 0. Second we need to know this special triangle: If you take the sine of 45, you see that it will be:\[\sin 45=\frac{ 1 }{ \sqrt{2} }=\frac{ 1 }{ \sqrt{2} }\frac{ \sqrt{2} }{ \sqrt{2} }=\frac{ \sqrt{2} }{ 2 }\]So,\[\sin^{-1} (-\frac{ \sqrt{2} }{ 2 })=-45°=-\frac{ \pi }{ 4 }\]

Answer 8

Sorry, I made a mistake on the first problem, it is actually -pi/6 because the sine wave isn't uniform.