Finding the inverse of Sin

  • #1

Homework Statement



This is from a physics problem, but my question is more mathematically oriented.

After working through the problem, I arrive at the last step.

Sin(x)=.967

The question says that there are two possible angles for x.



The Attempt at a Solution



arcsin(.967) gives me 75.2°, which is one of the correct answers.
I am stumped when it comes to finding the second possible angle, which happens to be approximately 105°

I understand that the inverse trig functions are restricted, so how do I find all possible answers for these kinds of questions.

Thank you!
 

Answers and Replies

  • #2
392
17
Think of the unit circle. [itex]sin(\theta)[/itex] is the y-coordinate on the unit circle. So, [itex]0.967 = sin(\theta) > 0[/itex] in the first and second quadrants. If you draw a triangle inside the unit circle with angle of 75.2 degrees from the x-axis, note where [itex]sin(\theta)[/itex] is. Notice there is another point with that same height in quadrant 2. Simply taking the mirror of this triangle across the y-axis will give you that point. Now calculate the new angle formed from the x-axis!

The angle formed with the trinagle and the Y-AXIS is 14.8. The new angle will be 75.2 + 2*14.8 = 104.8
 
  • #3
tiny-tim
Science Advisor
Homework Helper
25,836
251
hi physicsdreams! :smile:
I understand that the inverse trig functions are restricted

yes, eg arcsinx is defined as having only one value, the "principal value" (between ±90°)

but the solutions to "sinx = 0.967" are arcsin0.967 + n.360° and 180° - arcsin0.967 + n.360° for any whole number n :wink:

(look at a graph of sin)
 
  • #4
Think of the unit circle. [itex]sin(\theta)[/itex] is the y-coordinate on the unit circle. So, [itex]0.967 = sin(\theta) > 0[/itex] in the first and second quadrants. If you draw a triangle inside the unit circle with angle of 75.2 degrees from the x-axis, note where [itex]sin(\theta)[/itex] is. Notice there is another point with that same height in quadrant 2. Simply taking the mirror of this triangle across the y-axis will give you that point. Now calculate the new angle formed from the x-axis!

The angle formed with the trinagle and the Y-AXIS is 14.8. The new angle will be 75.2 + 2*14.8 = 104.8

Alright, I think I can visuallize it now.
Basically, I take 180-(first angle) and I should get my second angle.
Thanks!

P.S. Do I do the same thing for the other trig functions (cos, tan)?
 
  • #5
392
17
Well, it's not always 180-[itex]\theta[/itex], that just happened to be true in this case. For questions like this that involve sine and cosine, just look at where the point in on the unit circle, find a different point on the circle that has the same sign value, and then find [itex]\theta[/itex] for that point.

Tangent is a little bit more tricky. Think about it geometrically in the unit circle. If [itex]tan(\theta)[/itex] is negative, what two quadrants can it possibly be in? If [itex]tan(\theta)[/itex] is positive, what two quadrants can it be in? Then [itex]tan(\theta)[/itex] = 0, undefined should be much simpler.
 

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