Sine rule and a percentage-error interval · source reference

Printed question

In triangle PQR, $\angle\mathrm{PRQ}=\frac{2\pi}3$, $\angle\mathrm{RPQ}=\frac\pi6+\theta$, and PR = $a$ units, where $a$ is a positive real constant.

(i) Show that $\mathrm{PQ}=\frac{\sqrt3a}{\cos\theta-\sqrt3\sin\theta}$. [4]

(ii) Given that $\theta$ is sufficiently small, show that $\mathrm{PQ}\approx\sqrt3a(1+b\theta+c\theta^2)$, where $b,c$ are real constants. [3]

(iii) Given that $\theta>0$, find the range of values of $\theta$ such that the percentage error of the approximation in (ii) is less than 5%. [2]

[2023/SAJC/Prelim/P2/2]

Printed working

$\angle\mathrm{PQR}=\pi-2\pi/3-(\pi/6+\theta)=\pi/6-\theta$.

Sine rule: $\mathrm{PQ}/\sin(2\pi/3)=a/\sin(\pi/6-\theta)$.

$\mathrm{PQ}=a(\sqrt3/2)/[(1/2)\cos\theta-(\sqrt3/2)\sin\theta]=\sqrt3a/(\cos\theta-\sqrt3\sin\theta)$.

$\mathrm{PQ}\approx\sqrt3a/[1-\theta^2/2-\sqrt3\theta]=\sqrt3a[1-(\sqrt3\theta+\theta^2/2)]^{-1}$.

$\approx\sqrt3a[1+(\sqrt3\theta+\theta^2/2)+(\sqrt3\theta+\theta^2/2)^2+\cdots]\approx\sqrt3a(1+\sqrt3\theta+7\theta^2/2)$.

$f(\theta)=|\text{approximate PQ}-\text{actual PQ}|/|\text{actual PQ}|$; after cancelling $\sqrt3a$, $f(\theta)=|(1+\sqrt3\theta+7\theta^2/2)(\cos\theta-\sqrt3\sin\theta)-1|$.

Solve $f(\theta)<0.05$ on the physical triangle interval. Printed GC result $0<\theta<0.194$ (3 s.f.).

Teacher note 1

From MF27 under Trigonometry, $\sin(A-B)=\sin A\cos B-\cos A\sin B$, replacing $A$ by $\pi/6$ and $B$ by $\theta$.

Teacher note 2

From MF27 use Maclaurin expansion for $\cos\theta$ and $\sin\theta$. Then apply $(1-x)^{-1}=1+x+x^2+\cdots$, replacing $x$ by $\sqrt3\theta+\theta^2/2$.

Teacher note 3

Percentage error $=|\text{approximate value}-\text{actual value}|/|\text{actual value}|\times100=100f(\theta)$.

Diagram 09-18-triangle

Source triangle labels P,Q,R,a,2π/3,π/6+θ retained; schematic a=1.

Diagram 09-18-error

Mathematical review

(i) $\angle PQR=\pi/6-\theta$. By the sine rule, $PQ/a=\sin(2\pi/3)/\sin(\pi/6-\theta)$, giving the stated formula.

(ii) The denominator is $1-\sqrt3\theta-\theta^2/2+O(\theta^3)$. Its reciprocal is $1+\sqrt3\theta+7\theta^2/2+O(\theta^3)$; hence $b=\sqrt3,c=7/2$.

(iii) On the physical interval $0<\theta<\pi/6$, the relative error is $|(1+\sqrt3\theta+7\theta^2/2)(\cos\theta-\sqrt3\sin\theta)-1|$. To verify a unique accuracy interval, put $J=(1+\sqrt3\theta+7\theta^2/2)(\cos\theta-\sqrt3\sin\theta)$. On $0<\theta<\pi/6$, $J>0$ and $J\prime/\cos\theta=4\theta-(7\sqrt3/2)\theta^2-(4+8\sqrt3\theta+7\theta^2/2)\tan\theta<0$, since $\tan\theta>\theta$. As $J(0)=1$, the error $1-J$ strictly increases. Numerically it reaches $0.05$ at $\theta\approx0.194141$ radians and increases to that crossing. Thus $0<\theta<0.194141$ (approximately $0.194$ at 3 s.f.). The factor $a$ cancels from the percentage error.

Provenance

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