What are Small Angle Approximations?

The small angle approximations, as given in the Edexcel Formula Booklet, are:


\cos(\theta)\approx 1-\frac{\theta^2}{2}


These approximations can only be used when \theta is small. Hence why we call them ‘small angle’ approximations. Furthermore, \theta must be measured in radians.

Here we can see each of the trigonometric graphs plotted against their given approximations. We can see from these plots, the values of \theta for which these approximations are good.  Evidently, \sin(\theta)\approx \theta and \tan(\theta)\approx\theta appear to be good approximations for -\frac{\pi}{4}\leq\theta\leq\frac{\pi}{4} whereas \cos(\theta)\approx 1-\frac{\theta^2}{2} works well for -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}.

What about larger angle values?

Functions can be approximated by polynomials by taking a Taylor Expansion (what is a Taylor Expansion?). These expansions are, of course, dependent on where we are taking the approximation. Recall that for the small angle approximations above, they are only valid for values of \theta around zero. For other ranges, the Taylor approximation will look different. Note that when the Taylor series is taken for values around zero, we call it a Maclaurin series.

It can be shown that the Maclaurin series for \sin(\theta) and \cos(\theta) (\tan(\theta) isn’t as simple) are given by:


\cos(\theta)\approx 1-\frac{\theta^2}{2}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}+...

Notice that \sin(\theta) has only odd powers of \theta whereas \cos(\theta) has only even powers. As we can see, both expansions have alternating signs and increasing powers of \theta. Hence, the more terms included in the expansion, the larger the range of \theta values the expansion will be a good approximation for. This explains why the small angle approximation for \cos(\theta) works for a larger range of \theta values – it is quadratic whereas the approximations for \sin(\theta) and \tan(\theta) are linear.


Find the percentage error in the small angle approximation for \sin(\theta) when \theta=0.22 radians.


Firstly, the small angle approximation says that \sin(0.22)\approx 0.22. However, the actual value is \sin(0.22)=0.21823 to 5 decimal places. Hence, the error is roughly \vert 0.21823-0.22\vert=0.00177. Finally, dividing by the true value and multiplying by 100 gives the percentage error as approximately 0.805%. 

Find the approximate value of \frac{1-\cos(3\theta)}{\theta\tan(2\theta)} for small angles.


Using the small angle approximations:

\frac{1-\cos(3\theta)}{\theta\tan(2\theta)}\approx \frac{1-\left(1-\frac{(3\theta)^2}{2}\right)}{\theta\times 2\theta}=\frac{\frac{9\theta^2}{2}}{2\theta^2}=\frac{9}{4}

Hence, for \frac{1-\cos(3\theta)}{\theta\tan(2\theta)} is approximately \frac{9}{4} for \theta close to 0.

  1. Show that for values of \theta (in radians) close to zero, \frac{5\cos(2\theta)-\sin(\theta)-2}{1-\tan(2\theta)}\approx 3+5\theta.
  2. Hence state the approximate value of \frac{5\cos(2\theta)-\sin(\theta)-2}{1-\tan(2\theta)} for small angles.


  1. Using small angle approximations: \begin{array}{lll}\frac{5\cos(2\theta)-\sin(\theta)-2}{1-\tan(2\theta)}&\approx&\frac{5\left(1-\frac{(2\theta)^2}{2}\right)-\theta-2}{1-2\theta}=\frac{3-10\theta^2-\theta}{1-2\theta}\\&=&\frac{(1-2\theta)(3+5\theta)}{1-2\theta}=3+5\theta\end{array} 
  2. It follows from part 1, that for small values of \theta, 5\theta is also approximately 0 and so 3+5\theta\approx 3. Hence, \frac{5\cos(2\theta)-\sin(\theta)-2}{1-\tan(2\theta)}\approx 3.