Use linear interpolation to estimate sin 36 using as your 'known' values 0 & 60?

In summary, linear interpolation is a method for estimating values between two known data points by creating a straight line between them and using its equation to calculate values at specific points. It is commonly used to estimate sine values because it is simple and accurate, especially for trigonometric functions without specific values listed. To use linear interpolation for sine values, two known points and their corresponding sine values are needed. However, it has limitations such as assuming linear data and decreasing accuracy with greater distance between known points. Furthermore, it cannot be used to estimate values outside of the given range.
  • #1
escobar147
31
0
here is the answer:

36/60 = x/.8660

60x = .8660 x 36

60x = 31.176

x = .5196

however, where does the value: 0.8660 initially come from?

any help would be appreciated
 
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  • #2
escobar147 said:
however, where does the value: 0.8660 initially come from?
sin(60°) ~= 0.8660 (and sin(0°) = 0).
 
  • #3
rcgldr said:
sin(60°) ~= 0.8660 (and sin(0°) = 0).

thanks
 

1. How does linear interpolation work?

Linear interpolation is a method for estimating values between two known data points. It involves creating a straight line between the two points and using the equation of that line to calculate the value at a specific point along the line.

2. Why is linear interpolation used to estimate sine values?

Linear interpolation is commonly used to estimate sine values because it is a simple and accurate method for approximating values between two known points. This is especially useful for trigonometric functions like sine, which may not have specific values listed in a table or calculator.

3. How do you use linear interpolation to estimate sine values?

To estimate sine values using linear interpolation, you need to have two known values and their corresponding sine values. Then, you can find the slope of the line between these two points and use the equation of a line (y=mx+b) to calculate the value at a specific angle along the line.

4. What are the limitations of using linear interpolation to estimate sine values?

Linear interpolation is only accurate for estimating values between two known points. It assumes that the data is linear, which may not always be the case. Additionally, the accuracy of the estimation may decrease as the distance between the known points increases.

5. Can linear interpolation be used to estimate sine values for angles outside of the given range?

No, linear interpolation can only be used to estimate values within the range of the two known points. Estimating values outside of this range may result in inaccurate or unreliable estimations.

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