Small-angle series in a triangle · source reference

Printed question

(i) Show that the first two non-zero terms of the Maclaurin series for $\tan\theta$ is given by $\theta+\frac13\theta^3$. You may use the standard results given in the List of Formulae (MF26). [2]

In the right-angled triangle OBC shown above, point A lies on OB such that OA = 1, OB = $x$, where $x>1$ and OC = 1. It is given that angle COB is $\frac\pi2$ radians and that angle ACB is $\theta$ radians (see diagram).

(ii) Show that $\mathrm{AB}=\frac{2\tan\theta}{1-\tan\theta}$. [2]

(iii) Given that $\theta$ is a sufficiently small angle, show that $\mathrm{AB}\approx a\theta+b\theta^2+c\theta^3$ for exact real constants $a,b,c$ to be determined. [3]

[2022/ASRJC/Prelim/I/4]

Printed working

$\tan\theta=\sin\theta(\cos\theta)^{-1}\approx(\theta-\theta^3/3!)(1-\theta^2/2!)^{-1}\approx(\theta-\theta^3/3!)(1+\theta^2/2)=\theta+\theta^3/2-\theta^3/6=\theta+\theta^3/3$.

$x=\tan(\pi/4+\theta)=\frac{\tan(\pi/4)+\tan\theta}{1-\tan(\pi/4)\tan\theta}=\frac{1+\tan\theta}{1-\tan\theta}$.

$\mathrm{AB}=x-1=2\tan\theta/(1-\tan\theta)$.

$\mathrm{AB}\approx2(\theta+\theta^3/3)[1-(\theta+\theta^3/3)]^{-1}\approx2(\theta+\theta^3/3)[1+(\theta+\theta^3/3)+(\theta+\theta^3/3)^2]$.

$\approx(2\theta+2\theta^3/3)(1+\theta+\theta^2)=2\theta+2\theta^2+8\theta^3/3$. $a=2,b=2,c=8/3$.

Teacher note 1

Maclaurin expansions for $\sin\theta$ and $\cos\theta$ can be found in MF27. Apply $(1-x)^{-1}=1+x+x^2+x^3+\cdots$, replacing $x$ by $\theta^2/2$.

Teacher note 2

From MF27 under Trigonometry, $\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$. Replace $A$ by $\pi/4$ and $B$ by $\theta$.

Teacher note 3

Use the result of (i) for $\tan\theta$, then apply $(1-x)^{-1}=1+x+x^2+x^3+\cdots$, replacing $x$ by $\theta+\theta^3/3$.

Diagram 09-17-triangle

OC=OA=1; OB=x>1; θ=angle ACB; right angle COB; example θ=0.2 only.

Mathematical review

(i) Divide $\sin\theta=\theta-\theta^3/6+\cdots$ by $\cos\theta=1-\theta^2/2+\cdots$: $\tan\theta=\theta+\theta^3/3+O(\theta^5)$.

(ii) $\tan(\pi/4+\theta)=(1+\tan\theta)/(1-\tan\theta)=x$, and $AB=x-1=2\tan\theta/(1-\tan\theta)$.

(iii) Let $u=\tan\theta$. Then $2u/(1-u)=2(u+u^2+u^3)+O(u^4)=2\theta+2\theta^2+8\theta^3/3+O(\theta^4)$. Thus $a=2,b=2,c=8/3$. Geometrically $0<\theta<\pi/4$; the polynomial approximation requires a sufficiently small angle in radians.

Provenance

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