Locate the crossing before taking the absolute area · source reference

Printed question

(i) Verify without a calculator that $\theta=\pi/4$ is a root of $\tan^2\theta-2\cos^2\theta=0$. [1]

(ii) By sketching $y=\tan^2(x/2)$ and $y=2\cos^2(x/2)$, solve exactly $\tan^2(x/2)>2\cos^2(x/2)$ for $-\pi<x<\pi$. [2]

(iii) Hence evaluate exactly $\int_0^{2\pi/3}|\tan^2(x/2)-2\cos^2(x/2)|dx$. [6]

[2011/IJC/Prelim/I/11]

Printed working

(i)

$$\tan^2\frac\pi4-2\cos^2\frac\pi4=1-2(1/\sqrt2)^2=0.$$

Therefore $\theta=\pi/4$ is a root.

(ii)

From the diagram, $\tan^2(x/2)>2\cos^2(x/2)$ implies $-\pi<x<-\pi/2$ or $\pi/2<x<\pi$.

(iii)

$$\begin{aligned}\int_0^{2\pi/3}|\tan^2(x/2)-2\cos^2(x/2)|dx&=-\int_0^{\pi/2}[\tan^2(x/2)-2\cos^2(x/2)]dx+\int_{\pi/2}^{2\pi/3}[\tan^2(x/2)-2\cos^2(x/2)]dx\\&=-\int_0^{\pi/2}[\sec^2(x/2)-1-(1+\cos x)]dx+\int_{\pi/2}^{2\pi/3}[\sec^2(x/2)-1-(1+\cos x)]dx\\&=-[2\tan(x/2)-2x-\sin x]_0^{\pi/2}+[2\tan(x/2)-2x-\sin x]_{\pi/2}^{2\pi/3}\\&=-(2-\pi-1)+[(2\sqrt3-4\pi/3-\sqrt3/2)-(2-\pi-1)]\\&=\frac{3\sqrt3}2+\frac{2\pi}3-2.\end{aligned}$$

Teacher notes

The vertical lines $x=-\pi$ and $x=\pi$ are the vertical asymptotes of $y=\tan^2(x/2)$.

For $0\le x\le\pi/2$, $\tan^2(x/2)-2\cos^2(x/2)\le0$; for $\pi/2<x\le2\pi/3$ it is $>0$.

Used results: (1) From MF27, $\cos2A=2\cos^2A-1$, replacing $A$ by $x/2$. (2) Not in MF27, $\sec^2A=1+\tan^2A$, replacing $A$ by $x/2$.

Printed answer key

(ii) $-\pi<x<-\pi/2$ or $\pi/2<x<\pi$ and the labelled native graph; (iii) $3\sqrt3/2+2\pi/3-2$. No separate (i) entry is printed.

Source diagrams in native form