### 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.

### 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.

Back to June Revision Exercise 3

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

Back to June Revision Exercise 3

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

Back to June Revision Exercise 3

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

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

Back to June Revision Exercise 3

### 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)
$\pi \int_4^9 y^2 ~dx$

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

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

$= 13.6$

Back to June Revision Exercise 3

### 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)

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

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

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

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

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

$= \frac{59}{180}$

Back to June Revision Exercise 3

### June Revision Exercise 4 Q7

(i)

$f(x) = \frac{4}{(1-3)\sqrt{x^2+4}}$

$= 4(1-3x)^{-1} (4)^{-\frac{1}{2}} (1 + \frac{x^2}{4})^{-\frac{1}{2}}$

$= 2[1 + 3x + (3x)^2 + (3x)^3 + \ldots][1 - \frac{1}{2} \frac{x^2}{4} + \ldots]$

$= 2[1 + 3x + (-\frac{1}{8}+9)x^2 + (-\frac{3}{8}+27)x^3 + \ldots]$

$= 2 + 6x + \frac{71}{4}x^2 + \frac{213}{4}x^2 + \ldots$

$|3x| \textless 1 \text{ and } |\frac{x^2}{4}| \textless 1$

$\Rightarrow \text{Range of values} = (-\frac{1}{3}, \frac{1}{3})$

$2 \mathrm{tan} y = f(x)$

$2 \mathrm{sec}^2 y \frac{dy}{dx} = f'(x)$

$2[2\mathrm{sec}^2 y \mathrm{tan}y (\frac{dy}{dx})^2 + \mathrm{sec}^2 y \frac{d^2y}{dx^2}] = f''(x)$

When $x = 0, f(0) = 2, f'(0)=6, f''(0)=\frac{71}{2}$

$2 \mathrm{tan}y=2 \Rightarrow y= \frac{\pi}{4}$

$2(2)\frac{dy}{dx}=6 \Rightarrow \frac{dy}{dx}=\frac{3}{2}$

$2[2(2)(1)(\frac{3}{2})^2 + 2 \frac{d^2y}{dx^2}]= \frac{71}{2} \Rightarrow = \frac{35}{8}$

Hence $y = \frac{\pi}{4} + \frac{3}{2}x + \frac{35}{8}\frac{x^2}{2!}+ \ldots = \frac{\pi}{4} + \frac{3}{2}x + \frac{35}{16}x^2 + \ldots$