CAIE A-Level Further Math AS 1.5 Polar Coordinates Questions

$x^{2}-x+\frac{1}{4}+y^{2}=\frac{1}{4} \Rightarrow r^{2}-r \cos \theta+\frac{1}{4}=\frac{1}{4}$
Uses $x^{2}+y^{2}=r^{2}$ and $x=r \cos \theta$.

$r(r-\cos \theta)=0$
Factorises.

$[r \neq 0 \Rightarrow] r=\cos \theta$
AG.

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$\sin 2 \theta=\cos \theta \Rightarrow 2 \sin \theta \cos \theta=\cos \theta$
Sets r values equal and uses $\sin 2 \theta=2 \sin \theta \cos \theta$.

$\cos \theta \neq 0 \Rightarrow \sin \theta=\frac{1}{2}$$\cos \theta \neq 0$ must be recognised.

$\left(\frac{1}{2} \sqrt{3}, \frac{1}{6} \pi\right)$
3

Initial line drawn and one curve correct.
Other curve correct.

Intersection marked in correct position and both
curves labelled.

3

$\frac{1}{2} \int_{0}^{\frac{1}{6} \pi} \sin ^{2} 2 \theta \mathrm{~d} \theta+\frac{1}{2} \int_{\frac{1}{6} \pi}^{\frac{1}{2} \pi} \cos ^{2} \theta \mathrm{~d} \theta$
Uses $\frac{1}{2} \int r^{2} \mathrm{~d} \theta$ with correct limits.

$\frac{1}{2} \int_{0}^{\frac{1}{6} \pi} \sin ^{2} 2 \theta \mathrm{~d} \theta=\frac{1}{4} \int_{0}^{\frac{1}{6} \pi} 1-\cos 4 \theta \mathrm{~d} \theta$
Integrates $\sin ^{2} 2 \theta$ using identity.

$=\frac{1}{4}\left[\theta-\frac{1}{4} \sin 4 \theta\right]_{0}^{\frac{1}{6} \pi}$$\frac{1}{2} \int_{\frac{1}{6} \pi}^{\frac{1}{2} \pi} \cos ^{2} \theta \mathrm{~d} \theta=\frac{1}{4} \int_{\frac{1}{6} \pi}^{\frac{1}{2} \pi} 1+\cos 2 \theta \mathrm{~d} \theta$
Integrates $\cos ^{2} \theta$ using identity.

$=\frac{1}{4}\left[\theta+\frac{1}{2} \sin 2 \theta\right]_{\frac{1}{6} \pi}^{\frac{1}{2} \pi}$$\frac{1}{4}\left(\frac{1}{6} \pi-\frac{1}{8} \sqrt{3}\right)+\frac{1}{4}\left(\frac{1}{2} \pi-\frac{1}{6} \pi-\frac{1}{4} \sqrt{3}\right)=\frac{1}{8}\left(\pi-\frac{3}{4} \sqrt{3}\right)$
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