Calculating sin-1(-1/2) by Hand

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In summary, the conversation discusses the process of finding the value of sin-1(-1/2) and the confusion surrounding the negative answer. It is determined that there are two possible answers on the interval [0, 2pi), and the principal value of sin-1 is the one nearest to 0. The conversation also touches on the concept of using sine as a y-coordinate of an angle.
  • #1
kg90
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sin-1(-1/2)


Can someone explain how to do this problem by hand. Thanks.
 
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  • #2
Welcome to PF!

kg90 said:
sin-1(-1/2)


Can someone explain how to do this problem by hand. Thanks.

Hi kg90! Welcome to PF! :smile:

May I check … do you know how to do sin-1(1/2)?
 
  • #3
What is your definition of sine? (Since a right triangle cannot have an angle whose sine is -1/2, I assume you are NOT using "opposite side divided by hypotenuse" as your definition.
 
  • #4
Hey thanks for replying. I do know how to do most problems like this. I just don't know how you get the negative answer for the problem I posted above. I always end up with a positive.

HallsofIvy, I'm not sure..
 
  • #5
Think about which quadrant you are working in and where sine, cosine, etc. are positive/negative?
 
  • #6
NoMoreExams said:
Think about which quadrant you are working in and where sine, cosine, etc. are positive/negative?

Yeah, but aren't there two quadrants where that would be true?
 
  • #7
Yes, and therefore on the interval [0, 2pi), there will be 2 answers.
 
  • #8
Think about it this way, you probably know or can convince yourself (and if you can't, try to) that f(x) = sin(x) crosses the x - axis twice on the interval [0, 2pi) (which points would that be?). Now how would you do that? Well you would say f(x) = sin(x) = 0, so I need to find x such that x = sin^{-1)(0). Now in your case you have g(x) = sin(x) - 1/2 and you are trying to find those x-intercepts, so you solve g(x) = sin(x) - 1/2 = 0 or in other words x = sin^{-1}(1/2).
 
  • #9
kg90 said:
Hey thanks for replying. I do know how to do most problems like this. I just don't know how you get the negative answer for the problem I posted above. I always end up with a positive.
NoMoreExams said:
Yes, and therefore on the interval [0, 2pi), there will be 2 answers.

Hi kg90! :smile:

(have a pi: π :smile:)

"sin-1" normally means the principal value of sin-1

that's the one in (-π/2,π/2] …

in other words, nearest to 0. :smile:

it may help to think of the sin as the y coordinate of the angle …
so sin-1(-1/2) is the angle where the y coordinate is -1/2 :wink:
 

1. How do I calculate sin-1(-1/2) by hand?

To calculate sin-1(-1/2) by hand, you will need to use the inverse sine function or arcsine function. This function is typically denoted as sin-1 or arcsin and is the inverse of the sine function. It is used to find the angle in radians that has a given sine value. In this case, we want to find the angle whose sine is -1/2.

2. What is the formula for calculating sin-1(-1/2) by hand?

The formula for calculating sin-1(-1/2) is sin-1(-1/2) = -π/6 + 2πn, where n is any integer. This formula can be derived from the unit circle, where the sine of -π/6 is equal to -1/2.

3. What steps do I need to follow to calculate sin-1(-1/2) by hand?

To calculate sin-1(-1/2) by hand, follow these steps:

  1. Draw a unit circle and label the points where the sine value is equal to -1/2.
  2. Identify the angle -π/6 on the unit circle and mark it as the reference angle for sin-1(-1/2).
  3. Use the formula sin-1(-1/2) = -π/6 + 2πn, where n is any integer, to find all possible angles that have a sine value of -1/2.
  4. Choose the angle that is between -π and π, as this is the range for inverse sine function.

4. Can I use a calculator to calculate sin-1(-1/2)?

Yes, most scientific calculators have a built-in inverse sine function, denoted as sin-1 or arcsin. You can simply enter -1/2 into the calculator and press the sin-1 or arcsin button to get the result.

5. Why is it important to know how to calculate sin-1(-1/2) by hand?

Knowing how to calculate sin-1(-1/2) by hand is important because it allows you to understand the concept of inverse trigonometric functions and their applications. It also helps you to verify the results obtained from a calculator and improves your problem-solving skills in mathematics and science.

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