Small angle approximationsAQA A-Level Maths: Revision notes

When $\theta$ is small and measured in radians: $\sin\theta\approx\theta,\qquad \tan\theta\approx\theta,\qquad \cos\theta\approx1-\frac{\theta^2}{2}.$ They work because, for a small angle, the arc, the opposite side and the tangent length in a right-angled triangle are almost equal. Check with $\theta=0.1$: $\sin0.1=0.0998$, $\tan0.1=0.1003$, $\cos0.1=0.9950$ and $1-\frac{0.1^2}{2}=0.995$. For small positive $\theta$, $\sin\theta<\theta<\tan\theta$, and all three are very close.

The approximations apply to whatever angle is inside the function, provided that angle is small. So $\sin3\theta\approx3\theta,\qquad\tan2\theta\approx2\theta,\qquad\cos4\theta\approx1-\frac{(4\theta)^2}{2}=1-8\theta^2,\qquad\sin\frac{\theta}{2}\approx\frac{\theta}{2}.$ Square the whole argument, including its coefficient. For example $\cos2\theta\approx1-\frac{(2\theta)^2}{2}=1-2\theta^2$, so $1-\cos2\theta\approx2\theta^2$.

To simplify a ratio or product, replace each trig function with its approximation, then cancel powers of $\theta$. Example: $\frac{1-\cos4\theta}{\theta\tan2\theta}\approx\frac{8\theta^2}{\theta\times2\theta}=4.$ Use $1-\cos k\theta\approx\frac{k^2\theta^2}{2}$ as a shortcut. Terms of higher power, such as $\theta^3$, are ignored when a lower power of $\theta$ is present, because they are far smaller.

Replace the trig functions to turn the equation into a polynomial, then solve and keep only the small root. Example: solve $\sin\theta+\cos\theta=1.02$. $\theta+1-\frac{\theta^2}{2}=1.02\ \Rightarrow\ \theta^2-2\theta+0.04=0\ \Rightarrow\ \theta=1\pm\sqrt{0.96}.$ The root $\theta=1-\sqrt{0.96}=0.0202$ is small, so the approximation is valid. The other root $1.98$ is not small, so it is rejected.

The errors grow as $\theta$ grows. The percentage error is $\frac{\text{approximate}-\text{exact}}{\text{exact}}\times100$. For $\theta=0.03$ the estimate $\tan\theta\approx\theta$ is wrong by only about $0.03\%$. In applications, an angle of elevation $\theta$ to a tower of height $h$ gives a distance $d=\frac{h}{\tan\theta}\approx\frac{h}{\theta}$. The slant distance $L$ and the horizontal distance $d$ satisfy $d=L\cos\theta$, so $L-d=L(1-\cos\theta)\approx\frac12L\theta^2$, a second-order difference that is usually negligible. Convert degrees to radians before using the approximations: $3^\circ=\frac{\pi}{60}\approx0.0524$.