2022TrialShort answer

Q16 (a) (i)

2 marks

Question

Prove that
$\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}=i\tan\frac\theta2.$

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Solution

Let $t=\tan\frac{\theta}{2}$ .

\begin{aligned} \text{LHS}&=\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}\\ &=\frac{\frac{1-t^2}{1+t^2}+i\frac{2t}{1+t^2}-1}{\frac{1-t^2}{1+t^2}+i\frac{2t}{1+t^2}+1}\\ &=\frac{1-t^2+2ti-1-t^2}{1-t^2+2ti+1+t^2}\\ &=\frac{2ti-2t^2}{2+2ti}\\ &=\frac{t(i-t)}{1+ti}\\ &=\frac{ti\left(1-\frac{t}{i}\right)}{1+ti}\\ &=\frac{ti(1+ti)}{1+ti}\\ &=it\\ &=i\tan\frac{\theta}{2}. \end{aligned}

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Syllabus alignment

ProofComplex Numbers

The Nature of ProofUse formal proof language, contradiction, counter-examples and rigorous arguments involving numbers and inequalities.
Other representations of complex numbersUse Euler's formula and convert between Cartesian, polar and exponential forms to solve problems.

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