2019 H2 Math Paper 1 Question 9 Part (ib): the trigonometric approach

Show that $\displaystyle \frac{w-1}{w+1}$ can be expressed as $k \tan \frac{1}{2} \theta$, where $k$ is a complex number to be found.

In the video I introduced the “half-angle trick” using complex numbers in exponential/Euler form to solve the 2019 H2 Math Paper 1 Question 9 (ib). In this blog post I will illustrate a trigonometric approach.

The main key in a trigonometric approach involves the double angle formula: while the sine double angle is pretty straightforward $\sin 2A = 2 \sin A \cos A$, there are 3 different version of the cosine double angle formula $\cos 2A = \cos^2 A – \sin^2 A = 2 \cos^2 A – 1 = 1 – 2 \sin^2 A$. For this question (and in some other applications), the trick is to use the appropriate formula to eliminate any “1”s in the question. For example, in our working we will encounter $\cos \theta – 1$. If we let $\cos \theta = 1 – 2 \sin^2 \frac{\theta}{2}$, then $\cos \theta – 1 = 1 – 2 \sin^2 \frac{\theta}{2} – 1 = -2 \sin^2 \frac{\theta}{2}$, so the “$-1$” was eliminated. Similarly, when dealing with $\cos \theta + 1$, we want to use the other variant of the double angle formula to get $\cos \theta + 1 = 2 \cos^2 \frac{\theta}{2} – 1 + 1 = 2 \cos^2 \frac{\theta}{2}$.

The solution:

\begin{align}
\frac{w-1}{w+1}
&= \frac{\cos \theta + i \sin \theta – 1}{\cos \theta + i \sin \theta + 1} \\
&= \frac{1 – 2 \sin^2 \frac{\theta}{2} + i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} – 1}{2\cos^2 \frac{\theta}{2} – 1 + i 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} + 1} \\
&= \frac{- 2 \sin^2 \frac{\theta}{2} + 2i \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2} + 2i \sin \frac{\theta}{2} \cos \frac{\theta}{2}} \\
&= \frac{2 \sin \frac{\theta}{2} (- \sin \frac{\theta}{2} + i \cos \frac{\theta}{2})}{2 \cos \frac{\theta}{2} (cos \frac{\theta}{2} + i \sin \frac{\theta}{2})} \\
&= \tan \frac{\theta}{2} \frac{- \sin \frac{\theta}{2} + i \cos \frac{\theta}{2}}{cos \frac{\theta}{2} + i \sin \frac{\theta}{2}} \times \frac{\cos \frac{\theta}{2}-i\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-i\sin \frac{\theta}{2}} \\
&= \tan \frac{\theta}{2} \frac{-\sin \frac{\theta}{2} \cos \frac{\theta}{2} +i\sin^2 \frac{\theta}{2} + i \cos^2 \frac{\theta}{2} -i^2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{\cos^2 \frac{\theta}{2}+\sin^2 \frac{\theta}{2}} \\
&= \tan \frac{\theta}{2} \frac{-\sin \frac{\theta}{2} \cos \frac{\theta}{2} + \sin \frac{\theta}{2} \cos \frac{\theta}{2} +i(\sin^2 \frac{\theta}{2} +\cos^2 \frac{\theta}{2}) }{1} \\
&= \left ( \tan \frac{\theta}{2} \right) i
\end{align}

Hence $k= i$.

Q6b discussion

Continuing from your method (making sure to use a new letter), we have $$\frac{a^2-b^2-2abi}{a+bi} = 0 + ic$$ Simplifying gives us $a^2-b^2 - 2abi = -bc + aci$. It looks very unwieldy, but let's use our usual techniques (comparing real and imaginary parts). Our final aim is to find $w$ in terms of $a$ so let's find $b$ and $c$ in terms of $a$. Give it a try first, and you can check against my working below:
Is $\sqrt{x^2} = x$? The many different modulus functions.
What is $\sqrt{x^2}$? Most of us will intuitive say "$x$": after all, $\sqrt{9} = \sqrt{3^2} = \sqrt{3}$, for example. However, what is $\sqrt{(-3)^2}$?
It is not $-3$ and is in fact $\sqrt{(-3)^2} = \sqrt{9} = 3$. Hence $\sqrt{x^2} =x$ is only valid if $x$ is non-negative. If $x$ is negative, it turns out that $\sqrt{x^2} = -x$.
The reason for this stems from definition: the symbol $\sqrt{ \cdot}$ is defined to be the "positive square root" when there are actually two possible square roots to every positive real number (this is the reason the equation $x^2 = k$ has two solutions, $\pm \sqrt{k}$, for positive $k$).