What is the value of θ when the error in sinθ ≈ θ is approximately 5%?

[Gee Wiz]

Homework Statement

In order to simplify problems in physics, we often use various approximations. For example, when we investigate diffraction and interference patterns at small angles θ, we frequently approximate sinθ and tanθ by θ (in radians). Here you will calculate over what range these are reasonable approximations.

For θ= 43° this approximation has an error of almost exactly 10%:

θ = 43° = 0.75 radians

sinθ=0.682

|sinθ-θ| / |sinθ| ≈ 10%

For what value of θ (to the nearest degree) is the error in sinθ ≈ θ approximately 5%?

Homework Equations

|sinθ-θ| / |sinθ| ≈ 10%

The Attempt at a Solution

So I was not having much luck finding the approximate for 5% so i tried to work backwards with the .75 radians and 10%. I set it up like
|sin.75-.75| / |sin.75| ≈ 10%
which works...so i tried to use the CAS calculator to quick solve the other question and I had no such luck. I attempted to have it solve |sinx-x| / |sinx| = .1...and it gave me an answer of false. I did get the right answer...but I literally just plugged in numbers until it worked out. I was wondering first if there is a better way to solve this not using CAS and second why didn't the CAS system work?

 

[Chestermiller]

I would start with the Taylor series expansion for sinθ in terms of θ. I would retain only the first two terms. I would then divide this by θ. The first term would then be 1, and the second (quadratic term) would be the deviation from 1. I would set the second term equal to 1/10, or 1/20, and solve for θ.

 

[voko]

For this problem, it might just be easier to use the bisection method than be creative.

 

[haruspex]

Chestermiller's method is an excellent way to get started, and if you need more precision then you can refine it using voko's 'binary chop' approach.

why didn't the CAS system work?

Maybe it would help if you got rid of the absolute value symbols. You know which of sin θ and θ will be the larger in the region of interest.

 

[Chestermiller]

Some other possibilities

1. Retain 3 terms in the taylor series expansion and solve the resulting quadratic equation in θ² using the quadratic formula.
2. Solve sinθ=0.9θ or sinθ = 0.95θ using Newton's method, with a starting guess from my previous post.

Chet

 

[Gee Wiz]

Thanks you, I appreciate the help