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 $k= i$.

]]>The lockdown/circuit breaker sure has brought its shares of challenges in teaching and learning. Lessons are certainly slightly less effective without the physical interactive component: I can’t just look over students’ shoulder and check on their progress, and it’s definitely been harder to gauge how my explanations are faring without the feedback from non-verbal cues. Not to mention the occasional lag and technical difficulties.

When we first moved to home based learning I initially tweaked my home computer setup that I used to record youtube video and incorporated a whiteboard to show my working. My monitors were positioned differently and I was suffering from really bad neck strain after classes. After tinkering here and there I think I’ve arrived at a setup I’m pretty satisfied with. I’m looking to incorporate more chalk and blackboard vs markers and whiteboard (I’ve found chalk dust to be a lot more easily managed than marker fumes/marker ink and dust), and I could always improve the workflow in terms of how I access things on the computer to show during class. But yeah, let’s work hard together over this period!

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JC came and went, and then the army and university. I have mainly forgotten about those videos until I was preparing for my masters program. I tried to fill in the gaps in my mathematical knowledge as I was transitioning from an engineering degree to mathematics (having mainly been exposed to only calculus/analysis and combinatorics/graph theory during my undergraduate days). After a bit of research, I happened onto the same linear algebra videos on MIT OpenCourseWare from all those years. This was a time where HD videos were just starting to get popular and more and more universities are starting to put up their material online (I myself have just finished an online course on programming with Python on Udacity while procrastinating on my final year project). So the video, at least the technical aspects (like the resolution), were dated. But the content material was fantastic and Professor Gilbert Strang (the lecturer for the linear algebra course) immediately vaulted himself into my personal list of favorite professors for the way they taught and made you think about rather deep concepts.

So that period of time came and went as well, as with my masters program and I’m now some years into tutoring full time. It certainly piqued my interest when I saw Professor Strang bubble up my youtube feed a few days ago with a newly uploaded video from MIT OpenCourseWare. “Intro: A New Way to Start Linear Algebra”. I didn’t need to take much convincing to click it to check it out. Sure brings back those fond memories of learning for me. Professor Strang has aged considerably since I first saw him (it’s been 15 years or so!) but it’s nice to see him still healthy and sharp and engaging as ever. And it’s so motivational, to find that after so many years, he’s found a new way to approach the topic (when his original way was already pretty darn good!) that brings about further insight into the subject.

And that brought about a round of personal reflection on my side. As I get better at my craft I’ve also been dabbling into some side projects (this website being one of them, with recording youtube videos and working on an app the more recent ones) the past couple of years. The recent circuit breaker measures, while making lessons and teaching in general slightly more challenging, has also given me more time to devote to these projects. Let’s hope I’d eventually get around to even a fraction of what Prof Strang has been able to accomplish as I continually improve myself in this pursuit of mathematics/teaching.

Check out the Professor Strang’s lectures yourself here: https://ocw.mit.edu/2020-vision

]]>- it is faster,
- it uses a similar concept for both lines and planes with slight modifications.

The downsides, however, is that it involves a rather unwieldy projection vector formula $(\mathbf{a}\cdot\mathbf{\hat{b}}) \mathbf{\hat{b}}$ and that minor details make a huge difference to whether we get the correct answer (e.g. where are our vectors pointing, $\overrightarrow{AB}$ or $\overrightarrow{BA}$? Is our final projection vector $\overrightarrow{AF}$ or $\overrightarrow{FB}$?).

In this post I will offer an alternative method to find foot of perpendicular using earlier concepts that some students find easier to understand.

We will use the example of finding the foot of perpendicular, $F$ from $B(4,6,-1)$ to the line $$l: \mathbf{r} = \begin{pmatrix} -1 \\ 5 \\ -3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, \lambda \in \mathbb{R}.$$

There are two key ideas behind this approach:

- $F$ lies on the line: hence we can use the equation of the line
- $BF$ is perpendicular to the line: hence $\overrightarrow{BF}\cdot \mathbf{d} = 0$.

Since $F$ lies on the line, we can use the equation of the line and write $$\overrightarrow{OF} = \begin{pmatrix} -1+\lambda \\ 5-\lambda \\ -3+\lambda \end{pmatrix}$$ We next find $\overrightarrow{BF} = \overrightarrow{OF} – \overrightarrow{OB} = \begin{pmatrix} -5+\lambda \\ 1-\lambda \\ 2+\lambda \end{pmatrix}$.

Since $BF$ is perpendicular to the line, $\overrightarrow{BF}$ is perpendicular to the direction vector of the line. \begin{gather} \overrightarrow{BF} \cdot \mathbf{d} = 0 \\ \begin{pmatrix} -5+\lambda \\ -1-\lambda \\ -2+\lambda \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = 0 \\ -6+3\lambda = 0 \\ \lambda = 2 \end{gather} Substituting that back into the equation for $\overrightarrow{OF}$ gives us $\overrightarrow{OF} = \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix}$.

We will use the example of finding the foot of perpendicular, $F$ from $B(4,6,-1)$ to the plane $$p: \mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = 4.$$

There are two key ideas behind this approach:

- We will form the equation of the line $BF$, noting that it is parallel to the normal vector of the plane.
- The intersection point between the line $BF$ and the plane is the foot of perpendicular $F$.

To form an equation for the line $BF$, we note that $B$ lies on the line and the normal vector of the plane is the direction vector of the line. Thus $$l_{BF}: \mathbf{r} = \begin{pmatrix} 4 \\ 6 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \lambda \in \mathbb{R}.$$ Take note to not confuse the line $BF$ and the vector $\overrightarrow{BF}$.

$F$ is the point of intersection between $l_{BF}$ and the plane $p$. We thus substitute the equation of the line into the plane: \begin{gather} \begin{pmatrix} 4+\lambda \\ 6+\lambda \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = 4 \\ 10+2\lambda = 4 \\ \lambda = -3 \end{gather} Substituting that back into the equation of the line gives us $\overrightarrow{OF} = \begin{pmatrix} 1 \\ 3 \\ -1 \end{pmatrix}$.

]]> We have $\displaystyle \int \frac{1}{(1-x^2)(1+x^2)} \; \mathrm{d}x = t + C$. Since the denominator is so complicated, ** partial fractions** is the way to go.

$$ \frac{1}{(1-x)(1+x)(1+x^2)} = \frac{A}{1-x} + \frac{B}{1+x} + \frac{Cx+D}{1+x^2}.$$

Give it a go!

Rate of increase at a constant rate $A=1$: So we have rate of increase = 1.

Rate of increase at a decrease is proportional to fourth power of $x$ with $k=1$: So we have rate of decrease $= x^4$.

Combining, we have $\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t} = 1-x^4$. Factorizing it gives us the required result.

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:

We have already calculated that $\displaystyle \frac{(w^*)^2}{w} = \frac{a^2-b^2-2abi}{a+bi}$. Notice there is a division in our working, so we can further simplify (“rationalize” using conjugates)! Give it a try.

]]> Instead, we are given an equation of a curve $x^2 + (y-5)^2 = 25$. We want the ** volume** of the water in the bowl. Able to spot which topic can help us tackle that?

Find the equation of the line of intersection between $$ p_1: \mathrm{r} \cdot \begin{pmatrix} 1 \\ 0 \\ 5 \end{pmatrix} = 7, \quad p_2: \mathrm{r} \cdot \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = -3.$$

The two important things we need to find an equation for a line are (1) a * point * on the line and (2) the

The key observation here is that the direction vector of the line of intersection is perpendicular to the normal vectors of both planes (try to visualize it!). Thus, to find the direction vector of the line of intersection, we can use the **cross product**. Hence $$\mathrm{d} = \begin{pmatrix} 1 \\ 0 \\ 5 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} -5 \\ 6 \\ 1 \end{pmatrix}$$

To do this, we will first convert the equations of our planes into cartesian form

\begin{align} x+5z &=7 \\ x+y-z &= -3 \end{align}

Unfortunately, 2 equations with 3 unknowns is not something we commonly encounter or solve. The reason that we have more unknowns than equations is that there are actually many possible points we can find (infinitely many, in fact!). We don’t need that many points: we just need one to form our equation of the line. Thus, we can actually specify a value for one of the unknowns and proceed. For example, let $z=0$. Then our equations become

\begin{align} x & = 7 \\ x+y&=-3 \end{align}

which we can solve to get $x=7$ and $y=-10$, giving us the point $(7,-10,0)$. Letting $z=0$ was just an arbitrary choice, you can actually let one of $x,y$ or $z$ be your favorite number and proceed and you will be able to get a point on the line (different from $(7,-10,0)$, but equally valid nevertheless).

Combining the two parts (point and direction vector) gives us the equation of our line of intersection, $$l: \mathrm{r} = \begin{pmatrix} 7 \\ -10 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -5 \\ 6 \\ 1 \end{pmatrix} , \lambda \in \mathbb{R}.$$

You should double check that using your GC will give us the same result (and much faster!)

And while I can see why the vector form isn’t as useful for further applications because there are so many different direction vectors (infinite, in fact), and that the normal vector is a much better way to describe planes because it is unique (up to a scalar multiple), the equation $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$ just isn’t very intuitive and didn’t make a lot of sense. In today’s post I’d try my best to explain the origins of such an equation.

Understanding how the equation comes about boils down to knowing what $\mathbf{a}$ and $\mathbf{r}$ represent. $\mathbf{a}$ is the position vector of a fixed point $A$ that we already know lies on the plane. In forming an equation of an object, we will like to know what equation any other random points on the object satisfies. (For example, we say $y=2x+3$ is an equation of a line because we can check that $(0,3)$ satisfies the equation (and hence lies on the line) while $(1,4)$ does not). $\mathbf{r}$ is used to represent the position vector of a random point on our plane. We will call the point $R$.

So, in the picture above, we have fixed $A$, and randomly pick a few points which we named $R_1, R_2$ and $R_3$. It turns out that, regardless of which $R$ we take, we can form a direction vector $\overrightarrow{AR}$. And the normal vector of our plane, $\mathbf{n}$ is so special in the sense that $\overrightarrow{AR}$ is perpendicular to $\mathbf{n}$. Thus we have $\overrightarrow{AR} \cdot \mathbf{n} = 0$.

The rest boils down to some algebra manipulation. $\overrightarrow{AR} = \overrightarrow{OR}-\overrightarrow{OA}$ so $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n} = 0$. Expansion gives $\mathbf{r} \cdot \mathbf{n} – \mathbf{a} \cdot \mathbf{n} = 0$ which leads us to our familiar $\mathbf{r}\cdot\mathbf{n} = \mathbf{a}\cdot\mathbf{n}$.

Hopefully this proof and explanation helps, but I bet that even then the equation still looks a bit weird (it did to me years ago too!). The trick is to simply keep practicing with it and it’d become second nature in no time!

]]>Let $a$ be any number. Let’s start with $a \cdot 0$. By the property of $0$, we have $a \cdot 0 = a \cdot (0 + 0)$. Using the distributive property gives us $a \cdot 0 + a \cdot 0$. Combining it all together we have $a\cdot 0 = a\cdot 0 + a\cdot 0$. But the definition of $0$ is such that it doesn’t change a number after addition. So $a \cdot 0 = 0$.

Note the meaning of $-a$ is the additive inverse of $a$, meaning that $a + (-a) = 0$ while the meaning of $-(ab)$ is the additive inverse of $ab$. Let’s add $ab$ with $(-a)b$.

$ab + (-a)b = (a + (-a)) \cdot b$ by the commutative and distributive laws. Since $-a$ is the additive inverse of $a$, this gives us $0 \cdot b$ which is $0$ from MP1.1. Thus $(-a)b$ is the additive inverse of $ab$ which is what we want to prove.

$ (-a)(-b) + -(ab) = (-a)(-b) + (-a)(b)$ by the MP1.2. By the distributive property, $(-a)(-b) + (-a)b = (-a)(-b + b) = (-a)0 = 0$ by MP1.1. Since $-(ab)$ is the additive inverse of $ab$, $(-a)(-b)$ must have been $ab$.

Suppose there exist the multiplicative inverse of $0$, which we will denote by $0^{-1}$. Then $0 \cdot 0^{-1} = 1$. But by MP1.1, $0$ times anything is 0 so $0=1$. This brings us a whole host of problems: $1 + 1 = 0 + 0 = 0$. So no matter what we do, every number we try to create will all become 0. Hence allowing the multiplicative inverse of 0 means that we can only work with one number: not a very interesting proposition. Conversely, to have more than one number to work with, we cannot allow the multiplicative inverse of 0.

Case 1: $a=0$. That is part of the solution.

Case 2: $a \neq 0$. Then the multiplicative inverse $a^{-1}$ exists. So $ab= 0$ means $a^{-1} ab = a^{-1} 0 $ so $b=0$.

$ab (a^{-1}b^{-1}) = a a^{-1} b b^{-1}$ by the commutative and associative properties which gets us $1 \cdot 1 = 1$ by the property of the multiplicative identity $1$. Hence $a^{-1}b^{-1}$ is the multiplicative inverse of $ab$.

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