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

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In summary, the conversation is about finding all possible angles for a given trigonometric function. The problem involves solving for sin(x) = 0.967 and finding the two possible angles for x. The conversation explains how to use the unit circle to find the second angle, and also discusses the restrictions on inverse trigonometric functions. Finally, the conversation explores how to apply these concepts to other trigonometric functions such as cosine and tangent.
  • #1
physicsdreams
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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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  • #2
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
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)
 
  • #4
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)?
 
  • #5
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.
 

1. How do you find the inverse of Sin?

The inverse of Sin can be found by using the inverse sine function, also known as arcsine, which is denoted as sin^-1. This function will return an angle in radians that is the inverse of the sine of a given number.

2. What is the formula for finding the inverse of Sin?

The formula for finding the inverse of Sin is given by sin^-1(x) = arcsin(x) = y, where x is the input value and y is the output angle in radians.

3. Is the inverse of Sin the same as the reciprocal of Sin?

No, the inverse of Sin is not the same as the reciprocal of Sin. The reciprocal of Sin is equal to 1/sin(x), while the inverse of Sin is equal to arcsin(x).

4. What is the range of values for the inverse of Sin?

The range of values for the inverse of Sin is between -π/2 and π/2, inclusive. This means that the output angle will always be within this range, regardless of the input value.

5. How is the inverse of Sin used in real-life applications?

The inverse of Sin is commonly used in trigonometry and geometry to find missing angles in a triangle. It is also used in physics and engineering to calculate the angle of a force or vector, as well as in signal processing and computer graphics to manipulate angles and rotations.

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