Why can we approximate sin (x) as x? Over what range is the approximation valid? Does it also work in degrees?

Traditionally, we might think of sin(x) as the ratio of the opposite side to the hypotenuse in a right angled triangle of angle x. We therefore draw a diagram of the right angled triangle with sides o, a, h, and define sin(x) as the ratio o:h. Now we insert the angle x, and furthermore draw the circle of radius a from the point where the angle is measured. The arc length s contained within the triangle therefore has length ax, from the definition of the radian. We now notice that as x gets smaller and smaller, s and o become nearly equivalent. Therefore we say that o  ≈ s  ≈ ax and therefore that sin(x)  ≈ ax/h. Notice also that as x gets smaller, the triangle gets closer and closer to being isoceles and therefore we can claim that  ≈ a. It therefore follows that for small x, sin(x)  ≈ ax/a = x! Clearly our derivation only holds valid for small x. But how small? Numerical calculation gives an easy way to try this. Sin(0.5) and 0.5 have a percentage difference of just 4.29%. However, sin(1) and 1 have a difference of 20%. Clearly, it is appropriate to use only when x << 1. Finally, we now see that this derivation does not hold for degrees. The crucial step is that we measured the arc length by ax, whereas ind degrees this does not hold. We need to find how many degrees are in a radian.2pi r = 360 d, therefore 1 deg = pi/180 rad. Therefore s=ax*pi/180, and therefore sin(x)  ≈ pi/180 * x, not x.

Related Oxbridge Preparation Mentoring answers

All answers ▸

Interview: (Shown protein structure of virus capsid) Explain what you think this structure is, and how knowing the structure may be of relevance to medicine?


Cambridge Law Test: Who benefits from freedom of speech? Who loses?


Geography interview- Is terrorise or climate change a larger threat? (Real question I got asked)


What is the most important thing to do in the interview?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences