Interesting Trigonometry Question

We know how A’levels like to combine a few topics across. We also know how bad trigonometry can be. I was coaching my International Bachelorette (IB) class last week and came across a fairly interesting question. It tests students on their abilities to manage double angle identities. Not a tough question, but definitely good practice 🙂
FYI: All double angle identities are found in MF26.
Here it is…

(a) Show that $(1 + i \text{tan} \theta)^n + (1 - i \text{tan} \theta)^n = \frac{2 \text{cos} n \theta}{ \text{cos}^n \theta}$, $\text{cos} \theta \neq 0$.

(b)
(i) Use the double angle identity $\text{tan} 2 \theta = \frac{2 \text{tan} \theta}{1 - \text{tan}^2 \theta}$ to show that $\text{tan} \frac{\pi}{8} = \sqrt{2} - 1$.

(ii) Show that $\text{cos} 4x = 8 \text{cos}^4 x - 8 \text{cos}^2 x + 1$.

(iii) Hence find the value of $\int^{\frac{\pi}{8}}_0 {\frac{2 \text{cos} 4x}{ \text{cos}^2 x}}~dx$ exactly.

Integration By Parts

Today, I’ll share a little something about Integration by Parts. I want to share this because I observe that several students are over-reliant on the LIATE to perform Integration by Parts. This caused them to overlook/ appreciate its use.

Lets start by reviewing the formula for Integration by Parts

$\int v \frac{du}{dx} ~dx = uv - \int u \frac{dv}{dx} ~dx$

I like to share with students that Integration by Parts have two interesting facts.

1. It allows us to integrate expression that we cannot integrate. Eg. $\text{ln}x$ or inverse trigonometric functions. This is also closely related to how LIATE is established.

2. This is closely related to point 1. That is, instead of trying to integrate the expression, we are differentiating it instead. And that itself, is a very important aspect.

Next, I like to point out to students that LIATE is a general rule of thumb.
GENERAL – because it does not work 100% of the times. And today, I’ll use an example to illustrate how LIATE actually fails.

$\int \frac{te^t}{(t+1)^2} ~ dt$

Here we observe two terms $\frac{t}{(t+1)^2}$ and $e^t$. Going by LIATE, we should be differentiating $\frac{t}{(t+1)^2}$ and integrating $e^t$.

$\int t^2 e^{t} ~ dt$
$= \frac{t}{(t+1)^2} e^{t} - \int \frac{(1)(t+1)^2 - t(2)(t+1)}{(t+1)^4} e^t ~dt$

Now observe what happened here, after applying integration by parts, it got “worst”. We are stuck to integrating $\frac{(1)(t+1)^2 - t(2)(t+1)}{(t+1)^4}$ with an exponential… This should raise some question marks. But we did follow LIATE.

I’ll leave students to try out this on their own. And feel free to ask questions here. Have fun!

Random Questions from 2016 Prelims #9

HCI P1 Q2

Solve the inequality $\frac{2}{4(x+1)^2+1} > 1$

Hence find $\int_{-1}^{\frac{\sqrt{3}-2}{2}} |1 - \frac{2}{4(x+1)^2+1}| dx$, leaving your answer in exact form.

Random Questions from 2016 Prelims #7

VJC P1 Q9

(i) Sketch the graph with equation $x^2 +(y-r)^2 = r^2$, where $r >0$ and $y \le r$

A hemispherical bowl of fixed radius $r$ cm is filled with water. Water drains out from a hole at the bottom of the bowl at a constant rate. When the depth of water if $h$cm (where $h \le r$).

(ii) Use your graph in (i) to show that the volume of water in the bowl is given by $V = \frac{\pi h^2}{3} (3r-h)$.

(iii) Find the rate of decrease of the depth of water in the bowl, given that a full bowl of water would become empty in 24 s,

(iv) without any differentiation, determine the slowest rate at which the depth of water is decreasing.

June Revision Exercise 3 Q11

(i)

$\frac{dV}{dt} = 300 - kV, k > 0$

$\int \frac{1}{300 - kV} dV = \int 1 dt$

$-\frac{1}{k} \mathrm{ln}|300 - kV| = t + C$

$300 - kV = Ae^{-kt}$

When $t = 0, V = 0 \Rightarrow A = 300$

$\therefore V=\frac{300(1-e^{-kt})}{k}$

When $t = 20, V=4500 \Rightarrow 4500 = \frac{300(1-e^{-20t})}{k}$

Using GC, $k = 0.030293$

When $V = 6000, t = 30.7$

$\Rightarrow 30.7- 20 = 10.7$ time interval between the first and second alarm.

(iii)
As $t \rightarrow \infty, V \rightarrow 9903 m^3$ which is impossible as the canal has only a fixed capacity of $6000 m^3$. Thus, the model is not valid for large values of $t$

(iv)
Assume that the weather condition remain unchanged.

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June Revision Exercise 3 Q10

(i)

$\frac{d\theta}{dt} \propto (\theta - 4)$

$\frac{d\theta}{dt} = - k(\theta - 4), k > 0$

$\int \frac{1}{\theta - 4} ~d \theta = \int -k ~dt$

$\mathrm{ln}|\theta - 4| = -kt + C$

$\theta - 4 = \pm e^{-kt+C}$

$\theta = 4 + Ae^{-kt}$

Given $t = 0, \theta = -1, \Rightarrow A = 14$

$\therefore \theta = 4 - 14 e^{-kt}$

Given $t = 4, \theta = -6, \Rightarrow k = \frac{1}{4}\mathrm{ln}\frac{7}{5}$

$\therefore \theta = 4 - 14 e^{-\frac{t}{4}\mathrm{ln}\frac{7}{5}}$

When $\theta = 2, t \approx 23.13 \text{ hours}$

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June Revision Exercise 3 Q9

(i)

$y= vx^2$

$\frac{dy}{dx}=2vx + \frac{dv}{dx}x^2$

$x^2 \frac{dy}{dx}=2xy+y^2$

$\Rightarrow x^2[2vx+\frac{dv}{dx}x^2] = 2xy + y^2$

$x^2 v (2x) + \frac{dv}{dx}x^4 - 2xy - y^2 = 0$

$\frac{dv}{dx}x^4 - v^2 x^4 = 0$

$\frac{dv}{dx} - v^2 = 0$

(ii)

$\frac{dv}{dx} = v^2$

$\int v^2 ~dv = \int 1 ~dx$

$-\frac{1}{v} = x + C$

$-\frac{1}{\frac{y}{x^2}} = x + C$

$- \frac{x^2}{y}= x+C$

$y= \frac{-x^2}{x+C}$

(iii)

When $C=1, y=\frac{-x^2}{x+1}$

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June Revision Exercise 3 Q8

$z = (1+x^2) \frac{dy}{dx}$

$\frac{dz}{dx} = (1+x^2) \frac{d^2y}{dx^2} + 2x \frac{dy}{dx}$

$\Rightarrow (1+x^2)\frac{d^2y}{dx^2}= 2x(1-\frac{dy}{dx})$

$\frac{dz}{dx} = 2x$

$\int 1 ~dz = \int 2x ~dx$

$z = x^2 + C$

$(1+x^2)\frac{dy}{dx}= x^2 +C$

$\frac{dy}{dx} = \frac{x^2+C}{1+x^2}$

$y = \int 1 + \frac{C-1}{1+x^2} ~dx$

$y= x + (C-1) \mathrm{tan}^{-1}x + D$

When $x=0, y=-1 \Rightarrow d=-1$

When $x = 0, \frac{dy}{dx}=2 \Rightarrow C=2$

$\therefore y=x-1+\mathrm{tan}^{-1}x$

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June Revision Exercise 3 Q7

(a)

$\int x(\mathrm{ln}x)^2 ~dx$

$=\frac{x^2}{2}(\mathrm{ln}x)^2 - \int \frac{x^2}{2} 2(\mathrm{ln}x)\frac{1}{x} ~dx$

$=\frac{x^2}{2}(\mathrm{ln}x)^2 - \int x \mathrm{ln}x ~dx$

$=\frac{x^2}{2}(\mathrm{ln}x)^2 - \frac{x^2}{2} \mathrm{ln}x + \int \frac{x^2}{2}\frac{1}{x}~dx$

$=\frac{x^2}{2}(\mathrm{ln}x)^2 - \frac{x^2}{2} \mathrm{ln}x + \frac{x^2}{4} + C$

(b)

$\int \frac{1}{1-\mathrm{cos}2x} ~dx$

$= \int \frac{1}{2\mathrm{sin}^2x} ~dx$

$= \frac{2}{2} \int \mathrm{cosec}^2x~dx$

$= -\frac{1}{2} \mathrm{cot}x + C$

(c)
(i)

(ii)

$\int_{\mathrm{ln}2}^{\mathrm{ln}3} x ~dy$

$=\int_2^3 t^2 \frac{1}{t} ~dt$

$=\Big| \frac{t^2}{2} \Big|_2^3$

$= 2.5 \text{units}^2$

(iii)
$\pi \int_4^9 y^2 ~dx$

$=\pi \int_2^3 (\mathrm{ln}t)^2(2t) ~dt$

$=2\pi \int_2^3 t(\mathrm{ln}t)^2 ~dt$

$= 13.6$

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June Revision Exercise 3 Q6

(a)(i)

Consider $y=\sqrt{a^2 - x^2} \Rightarrow y^2+x^2=a^2$ is a circle with centre Origin and radius $a$.

Thus, required area is the area of a quadrant with radius $a$

$\text{Area} = \frac{1}{4} \pi a^2$

(ii)

$\int_0^a \frac{x^2}{\sqrt{a^2-x^2}} ~dx$

$= \int_0^2 x\frac{x}{\sqrt{a^2-x^2}} ~dx$

$= - x \sqrt{a^2-x^2} \Big|_0^a - \int_0^a -\sqrt{a^2-x^2} ~dx$

$= 0 + \frac{1}{4} \pi a^2$

$= \frac{1}{4} \pi a^2$

(b)

$u=3x+5 \Rightarrow \frac{du}{dx}=3$

$= \int_{-2}^{-\frac{5}{3}} (x-2)(3x+5)^3 ~dx$

$=\int_{-1}^0 (\frac{u-5}{3}-2)u^3 \frac{1}{4} ~du$

$=\frac{1}{9} \int_{-1}^0 u^4 - 11u^3 ~du$

$=\frac{1}{9} \Big| \frac{1}{5}u^5 - \frac{11}{4}u^4 \Big|_{-1}^0$

$= \frac{59}{180}$

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