How can I find all possible angles for a given sine value?

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Homework Help Overview

The discussion revolves around finding all possible angles for a given sine value, specifically when sin(x) = 0.967. The original poster expresses confusion regarding the second angle after determining one angle using arcsin.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the unit circle to visualize sine values and the concept of angles in different quadrants. There is mention of the principal value of inverse sine and how to derive additional angles from it.

Discussion Status

Several participants provide insights into the relationship between sine values and angles in the unit circle, with some suggesting methods to find the second angle. There is an exploration of the general approach for sine and cosine functions, while also raising questions about the tangent function.

Contextual Notes

Participants note the restrictions of inverse trigonometric functions and the importance of understanding the unit circle for determining angles. There is a recognition that the approach may vary for different trigonometric functions.

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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!
 
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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
 
hi physicsdreams! :smile:
physicsdreams said:
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)
 
scurty said:
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)?
 
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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