Category Archives: Area

February 2, 2025 · 7:46 am

Trigonometric Limits

$\lim\limits_{x \to 0}\frac{sin(x)}{x}=?$

Remember $cos(x)=\frac{OA}{OB}=\frac{OA}{1}$ , hence $OA=cos(x)$ and the co-ordinate of $A$ is $(cos(x), 0)$ .

$sin(x)=\frac{AB}{OB}=\frac{AB}{1}$ , hence $AB=sin(x)$ and the co-ordinate of $B$ is $(cos(x), sin(x))$

And from the definition of $tan(x)$ we know $D$ is the point $(1, tan(x))$

Consider the areas of triangle $OAB$ , sector $OBC$ , and triangle $OCD$ .

We know from inspection of the above diagram that

Area $OAB<$ Area $OCB<$ Area $OCD$

Which means,

$\frac{1}{2}b_1 h_1<\frac{1}{2}r^2x<\frac{1}{2}b_2 h_2$

We can ignore all of the halves.

$cos(x)sin(x)<x<(1)tan(x)$

Remember $tan(x)=\frac{sin(x)}{cos(x)}$

$cos(x)sin(x)<x<\frac{sin(x)}{cos(x)}$

Divide everything by $sin(x)$ (as we are in the first quadrant we know $sin(x)>0$ , so we don’t need to worry about the inequality)

$cos(x)<\frac{x}{sin(x)}<\frac{1}{cos(x)}$

Invert everything and change the direction of the inequalities)

$\frac{1}{cos(x)}>\frac{sin(x)}{x}>cos(x)$

I am going to rewrite it as follows

$cos(x)<\frac{sin(x)}{x}<\frac{1}{cos(x)}$

because I like to use less thans rather than greater thans.

Now what happens as $x$ tends to $0$ ?

$cos(0)=1$

$1<\frac{sin(x)}{x}<\frac{1}{1}$

Hence by the squeeze theorem $\lim\limits_{x \to 0}\frac{sin(x)}{x}=1$

Now we know this limit, we are going to use it to find $\lim\limits_{x \to 0}\frac{1-cos(x)}{x}$

Multiply by $\frac{1+cos(x)}{1+cos(x)}$

$\lim\limits_{x \to 0}\frac{1-cos(x)}{x}\times \frac{1+cos(x)}{1+cos(x)}$

$\lim\limits_{x \to 0}\frac{1-cos^2(x)}{x(cos(x)+1)}$

$\lim\limits_{x \to 0}\frac{sin^2(x)}{x(cos(x)+1)}$

$\lim\limits_{x \to 0}\frac{sin(x)}{x}\times sin(x)(cos(x)+1)}$

$\lim\limits_{x \to 0}\frac{sin(x)}{x}\times \lim\limits_{x \to 0}sin(x)(cos(x)+1)$

If we evaluate the limits,

$(1)(sin(0)(cos(0)+1)=1\times 0 \times 2=0$

Hence, $\lim\limits_{x \to 0}\frac{1-cos(x)}{x}=0$

In the next post we are going to use these limits to differentiate sine and cosine functions.

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Filed under Area, Area of Triangles (Sine), Calculus, Identities, Trigonometry, Year 12 Mathematical Methods

Tagged as trig limits, trigonometric limits

December 14, 2024 · 6:56 am

Volume and Surface Area of a Conical Frustrum

My best attempt at drawing a Frustrum in Geogebra.

We have a truncated cone,

(1) $\begin{equation*}V=\frac{1}{3}\pi R_2^2(h_1+h_2)-\frac{1}{3}\pi R_1^2h_1\end{equation*}$

We are unlikely to know $h_1$ . Can we get $h_1$ in terms that we do know (i.e. $R_1, R_2, h_2$ )?

Think of similar triangles

$\Delta ABC \sim \Delta ADE (AA)$

$R_1 \parallel R_2$

$\angle {C}=\angle{E}$ (Corresponding Angles in Parallel Lines)

$\angle {B}=\angle {D}$ (Corresponding Angles in Parallel Lines)

Therefore

$\begin{equation*}\frac{h_1}{R_1}=\frac{h_1+h_2}{R_2}\end{equation}$

Rearrange to make $h_1$ the subject.

$\begin{equation*}h_1=\frac{h_2R_1}{R_2-R_1}\end{equation}$

Substitute into equation $(1)$

$\begin{equation*}V=\frac{1}{3} \pi((R_2^2(\frac{h_2R_1}{R_2-R_1})+h_2)-R_1^2(\frac{h_2R_1}{R_2-R_1}))\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi (R_2^2h_2+\frac{R_2^2h_2R_1}{R_2-R_1}-\frac{R_1^2h_2R_1}{R_2-R_1})\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi (R_2^2h_2+\frac{R_2^2h_2R_1-R_1^2h_2R_1}{R_2-R_1})\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi (R_2^2h_2+\frac{h_2R_1}{R_2-R_1}(R_2^2-R_1^2))\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi h_2(R_2^2+\frac{R_1}{R_2-R_1}(R_2-R_1)(R_2+R_1))\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi h_2(R_2^2+R_1(R_2+R_1))\end{equation}$

$\begin{equation*}V=\frac{1}{3} \pi h_2(R_2^2+R_1 R_2+R_1^2)\end{equation}$

Now let’s think about the surface area.

The surface area of a cone is $A=\pi r^2+\pi rs$ where $s$ is the slant height of the cone.

Once again, we need to subtract the ‘missing’ part of the cone.

(2) $\begin{equation*}A=\pi R_2^2+\pi R_2(s_1+s_2)+ \pi R_1^2- \pi R_1s_1\end{equation*}$

We don’t need to subtract the circle of the top cone because it is the top of the frustrum, but we do need to add it on.

Using similar triangles again

$\begin{equation*}\frac{s_1}{R_1}=\frac{s_1+s_2}{R_2}\end{equation}$

$\begin{equation*}s_1=\frac{s_2R_1}{R_2-R_1}\end{equation}$

Substitute into equation $(2)$

$\begin{equation*}A=\pi(R_2^2+R_2(\frac{s_2R_1}{R_2-R_1}+s_2)+\pi R_1^2-R_1(\frac{s_2R_1}{R_2-R_1}))\end{equation}$

$\begin{equation*}A=\pi(R_2^2+R_2s_2+R_1^2+\frac{s_2R_1}{R_2-R_1}(R_2-R_1))\end{equation}$

$\begin{equation*}A=\pi(R_2^2+R_2s_2+R_1s_2+R_1^2)\end{equation}$

$\begin{equation*}A =\pi (R_1^2+s_2(R_1+R_2)+R_2^2)\end{equation}$

And hence the curved surface area is $\pi s_2(R_1+R_2)$ .

Tricky Area Ratio Question

A circle of radius $r$ , is inscribed within an isosceles triangle $ABC$ . $CA=CB=5r$ . Given that $\angle{ACB}$ is acute, find the ratio of the area of the circle to that of triangle $ABC$ .

Mathematics Specialist 3AB Question15 page 55

I came across this question while searching for an area of sectors and segments question.

Here’s a diagram

We know $AC, AB and BC$ are tangents to the circle. Because the triangle is isosceles, the distance from $A$ to the circle is the same as the distance from $B$ to the circle.

$CP$ is perpendicular to $AB$ (because the triangle is isosceles).

Because it is proportional, i.e. $r$ and $5r$ , we can let $r=1$

Let $h=CP$

(1) $\begin{equation*}h=\sqrt{25-a^2}\end{equation*}$

but $h$ also equals

(2) $\begin{equation*}h=1+\sqrt{(5-a)^2+1}\end{equation*}$

Set equation $1$ equal to equation $2$

$\begin{equation*}\sqrt{25-a^2}={1+\sqrt{(5-a)^2+1}\end{equation}$

Square both sides

$\begin{equation*}25-a^2=1+(5-a)^2+1+2\sqrt{(5-a)^2+1}\end{equation}$

Expand and simplify

$\begin{equation*}25-a^2=2+25-10a+a^2+2\sqrt{(5-a)^2+1}\end{equation}$

$\begin{equation*}0=2-10a+2a^2+2\sqrt{(5-a)^2+1}\end{equation}$

Divide by $2$

$\begin{equation*}0=1-5a+a^2+\sqrt{(5-a)^2+1}\end{equation}$

$\begin{equation*}-\sqrt{(5-a)^2+1}=a^2-5a+1\end{equation}$

Square both sides

$\begin{equation*}(5-a)^2+1=a^4-5a^3+a^2-5a^3+25a^2-5a+a^2-5a+1\end{equation}$

$\begin{equation*}25-10a+a^2+1=a^4-10a^3+27a^2-10a+1\end{equation}$

$\begin{equation*}0=a^4-10a^3+26a^2-25 \end{equation}$

I solved this using a graphics calculator

$a=-0.8434, a=1.3068, a=4.5367, a=5$

We can reject $a=-0.8434$ , $a=4.5367$ , and $a=5$ . If $a=5$ , there isn’t a triangle, and if $a=4.5367$ $\angle{ACB}$ is not acute.

Hence the area of triangle $ABC=a\times h$

$\begin{equation*}A_{\Delta}=1.3068\times\sqrt{25-1.3068^2}\end{equation}$

(3) $\begin{equation*}A_{\Delta}=6.3069\end{equation*}$

(4) $\begin{equation*}A_{\circ}=\pi\end{equation*}$

Hence, equation $4$ divided by equation $3$ is

$\begin{equation*}\frac{\pi}{6.3069}\approx 0.5\end{equation}$

Perhaps I approached this question in the wrong way. Is there an easier process?

Three Circles – Area Problem

This is question 5 from the UK Maths Trust Senior Challenge October 2023.

I have tackled this in three ways; using non-right trig to find the area, Heron’s Law, and the Shoelace Formula.

Method 1

Use the area of a triangle formula

Use the cosine rule to find cosθ.

Once we have cosθ, we can find sinθ.

Hence the area is,

Method 2

Use Heron’s law.

Heron’s law is a way of calculating area of a triangle from the lengths of the three sides of the triangle.

This is my preferred method – simple and direct.

Method 3

Shoelace formula (Gauss’s Area formula)

We need to allocate each of the vertices a co-ordinate.

The co-ordinates are listed in an anti-clockwise direction.

This is probably a bit over the top, but once you get the hang of it, it’s very easy.