Question

Give the measure of the angle in radians and degrees
arcsin (-square root of 2/2

Answer 1

@satellite73 please help

Answer 2

@ganeshie8 please help

Answer 3

@tkhunny please help

Answer 4

Do you have a "Unit Circle" to look at? You should be quite familiar with \(-\pi/4\). Why negative and not positive?

Answer 5

Where did you get -pi/4

Answer 6

Why did you use pi/4 and not something like 3pi/4?

Answer 7

Did you find a "Unit Circle", yet?

Answer 8

yes

Answer 9

For one, the arcsin function typically is defined only for \(-\pi/2\le arcsin(x)\le\pi/2\). This rules out quite a few options.

Answer 10

What does your unit circle tell you about the sine of \(-\pi/4\)?

Answer 11

Its 45 degrees and is (square root of 2/2, square root of 2/2

Answer 12

No, that's \(\pi/4\). We need \(-\pi/4\)

When you find it, it will say, \(\sin(-45º) = -\sqrt{2}/2\). This should look quite tantalizing in light of the original problem statement.

Answer 13

Would that be 7pi/4?

Answer 14

Well, that is the same place, but it is NOT the answer to the original problem statement.

This is correct:

\(sin(-\pi/4) = -\sqrt{2}/2\)
\(sin(7\pi/4) = -\sqrt{2}/2\)

However, and this may come as a surprise:

\(arcsin(-\sqrt{2}/2) = -\pi/4\) and is absolutely NOT \(7\pi/4\). It is a Range Issue. The arcsin function does not contain \(7\pi/4\) in its Range.

Answer 15

Ok I understand, if the range was from -infinity to infinity, 7pi/4, would be a valid answer, but since it isn't you just have to slap on the negative to pi/4

Answer 16

Not quite, but that may get you to the next problem.

Answer 17

thanks for helping