Trigonometry

[Gorzie150]

If sin(r)=w and -π/2<r<0, determine (in terms of r), smallest possible solution to the equation sin(x)=w.

 

[stapel]

If sin(r)=w and -π/2<r<0, determine (in terms of r), smallest possible solution to the equation sin(x)=w.

Hint: Use the shape of the sine wave to find the x-value in the quadrant which has the same "height" as the sine wave in the fourth quadrant.

 

[DrPhil]

If sin(r)=w and -π/2<r<0, determine (in terms of r), smallest possible(??) solution to the equation sin(x)=w.

This is a very poorly stated question!

The solution r is in the 4th quadrant, and r is a negative number in (-pi/2,0). If you subtract 2pi, that is also a solution. You can keep subtracting 2pi forever, so there is no "smallest possible solution."

If you want the smallest magnitude solution, that will be |r|.

If you want the smallest positive solution, that will be in the 3rd quadrant. Might that be the word used in the question, rather than possible? Another responder has discussed how to find the angle in the 3rd quadrant, by looking at the same "height" on a sine wave - also you can look at the unit circle.