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Nitya Nigam Readers who study physics may have encountered the small angle rule in their classes. It's an essential tool in simplifying equations, both at a high school level and beyond. But what is the rule? You're probably familiar with the basic trigonometric ratios: sine, cosine and tangent. The small angle rules states that, for small angles (big surprise), sin(x) ≈ x ≈ tan(x). This is a very powerful statement, and we'll look at some of its applications in later articles, but in this article, we'll look at a few different ways to prove this statement. Diagrams are sourced from this link. First, let's try a visual approach. The graph below shows x, sin(x) and tan(x) on the same axes. It's clear that the lines are very close to each other for small values of x. Another visual approach is to consider the geometric definitions of the trigonometric ratios, which involve the three sides (opposite, adjacent and hypotenuse) of a right-angled triangle. Consider the diagram below: As you probably remember, sine = opposite / hypotenuse = O/H. For small θ, the arc s is very close to O in length, so we can say s ≈ O. The formula for arc length is rθ, where r is the radius of the circle, which is essentially H. So if s ≈ O and s ≈ H θ, then O ≈ H θ, and θ ≈ O / H ≈ sin(θ). By similar logic, we can say that s ≈ A θ, so θ ≈ A / H ≈ tan(θ).
The next method of derivation involves a bit of calculus called L'Hopital's rule. Check out Long Him's article on limits if you haven't encountered the concept before. This rule allows you to find the limits of functions as they approach a point at which they are undefined. Formally stated, if lim(x --> 0) f(x) = 0 and lim(x --> 0) g(x) = 0, then f(x)/g(x) is undefined, but lim(x --> 0) [f(x)/g(x)] = lim(x --> 0) [f'(x)/g'(x)], where f'(x) is the derivative of f(x). In the context of the small angle rule, we are trying to show that lim(x --> 0) [sin(x)/x] = 1, as this would imply that sin(x) = x for small x. sin(x) differentiates to cos(x) and x differentiates to 1, so applying L'Hopital's rule gives us lim(x --> 0) [sin(x)/x] = lim(x --> 0) [cos(x)/1] = lim(x --> 0) [cos(0)] = 1, as required. Similar logic using tan(x) gives lim(x --> 0) [sec^2(x)/1] = lim(x --> 0) [sec^2(0)] = 1. Our final method involves Maclaurin series, which you may have read about in Malhar's article where he uses them to prove Euler's identity. The Maclaurin series for sin(x) is x + (x^3 / 3!) + (x^5 / 5!) + ... and for small values of x, it is valid to truncate the series after the first term, yielding sin(x) ≈ x once again. Similarly, the Maclaurin series for tan(x) is x + (x^3 / 3) + (2/15 x^5) + ... and we can truncate this to give tan(x) ≈ x as required. Hopefully you found this article helpful! Let us know in the comments if you would like to learn about how the small angle rule is applied.
1 Comment
24/1/2021 12:29:43 am
I don't think the last two methods are valid. You cannot differentiate sin x to cos x from first principles without somehow resolving the limit x->0 of sin x / x ; by using the differential, both in L'hopital's rule and in the construction of a Maclaurin series, you may get the correct answer for the limit, but I'm not sure if it is proof.
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