Category Archives: Trigonometry

December 16, 2024 · 9:19 am

Interesting Sum

S=\sum_{n=1}^\infty (tan^{-1}(\frac{2}{n^2})), find S.

I came across this sum in An Imaginary Tale by Nahin and I was fascinated.

Let tan(\alpha)=n+1 and tan(\beta)=n-1.

Remember
tan(\alpha-\beta)=\frac{tan(\alpha)-tan(\beta)}{1+tan(\alpha)tan(\beta)}
Hence,
tan(\alpha-\beta)=\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}
tan(\alpha-\beta)=\frac{2}{1+n^2-1}
tan(\alpha-\beta)=\frac{2}{n^2}
Therefore,
\alpha-\beta=tan^{-1}(\frac{2}{n^2})
and
\alpha=tan^{-1}(n+1) and \beta=tan^{-1}(n-1)

tan^{-1}(n+1)-tan^{-1}(n-1)=tan^{-1}(\frac{2}{n^2})

Which means,

    \begin{equation*}S=\sum_{n=1}^\infty(tan^{-1}(n+1)-tan^{-1}(n-1))\end{equation}

Let’s try a few partial sums

S_4=tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)+tan^{-1}(5)-tan^{-1}(3)

S_4=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(4)+tan^{-1}(5)

S_6=tan^{-1}(2)-tan^{-1}(0)+tan^{-1}(3)-tan^{-1}(1)+tan^{-1}(4)-tan^{-1}(2)+tan^{-1}(5)-tan^{-1}(3)+tan^{-1}(6)-tan^{-1}(4)+tan^{-1}(7)-tan^{-1}(5)

S_6=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(6)+tan^{-1}(7)

Hence, S_N=-tan^{-1}(0)+-tan^{-1}(1)+tan^{-1}(N)+tan^{-1}(N+1)

S_N=-\frac{\pi}{4}-0+tan^{-1}(N)+tan^{-1}(N+1)

What happens as N\rightarrow \infty ?

\lim\limits_{N\to \infty}\ S_N=-\frac{\pi}{4}+\frac{\pi}{2}+\frac{\pi}{2}=\frac{3\pi}{4}

Because we know tan(\frac{\pi}{2}) is undefined.

1 Comment

Filed under Identities, Interesting Mathematics, Puzzles, Sequences, Trigonometry

Tagged as , ,

November 26, 2024 · 7:07 am

Infinite Product Expansion of cos (x)

Remember

(1)   \begin{equation*}cos(x)=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+...\end{equation*}

We know that cos(x)=0 for odd integer multiples of \frac{\pi}{2}, i.e. \frac{\pi}{2}, \frac{3\pi}{2}, ..., which is \frac{(2n-1)\pi}{2} for n\neq 0

Hence,

    \begin{equation*}0=1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+...\end{equation}

for x=\frac{(2n-1)\pi}{2}, n>0

We can factorise our cos(x) expansion

    \begin{equation*}(1-\frac{x^2}{r_1})(1-\frac{x^2}{r_2})...\end{equation}

We know r_1=\frac{\pi}{2}, r_2=\frac{3\pi}{2}, ...

    \begin{equation*}cos(x)=(1-\frac{x^2}{(\frac{\pi}{2})^2})(1-\frac{x^2}{(\frac{3\pi}{2})^2})...(1-\frac{x^2}{(\frac{(2n-1)\pi}{2})^2})\end{equation}

    \begin{equation*}cos(x)=\Pi_{n=1}^{\infty}(1-\frac{x^2}{(\frac{(2n-1)\pi}{2})^2})\end{equation}

    \begin{equation*}cos(x)=\Pi_{n=1}^{\infty}(1-\frac{4x^2}{(2n-1)^2\pi^2})\end{equation}

1 Comment

Filed under Factorising, Identities, Infinite Product Expansion, Interesting Mathematics, Polynomials, Trigonometry

Tagged as ,

October 28, 2024 · 6:51 am

Related Rates Question

A boat is moving towards the beach line at 3 metres/minute. On the boat is a rotating light, revolving at 4 revolutions/minute clockwise, as observed from the beach. There is a long straight wall on the beach line, as the boat approaches the beach, the light moves along the wall. Let x equal the displacement of the light from the point O on the wall, which faces the boat directly. See the diagram below.
Determine the velocity, in metres/minute, of the light when x=5 metres, and the distance of the boat from the beach D is 12 metres.

Mathematics Specialist Semester 2 Exam 2018


The light is rotating at 4 revolutions/minute, which means

    \begin{equation*}\frac{d\theta}{dt}=4\times\pi\end{equation}

We want to find \frac{dx}{dt} and we know \frac{d\theta}{dt} and \frac{dD}{dt}.

We need to find an equation connecting x, \theta, and D.

    \begin{equation*}tan (\theta)=\frac{x}{D}\end{equation}

Differentiate (implicitly) with respect to time.

    \begin{equation*}sec^2(\theta)\frac{d\theta}{dt}=\frac{\frac{dx}{dt}D-\frac{dD}{dt}x}{D^2}\end{equation}

Now we know x=5, and D=12, using pythagoras we can calulate the hypotenuse.


h=\sqrt{12^2+3^2}=13
sec(\theta)=\frac{13}{12}

    \begin{equation*}sec^2(\theta)\frac{d\theta}{dt}=\frac{\frac{dx}{dt}D-\frac{dD}{dt}x}{D^2}\end{equation}

    \begin{equation*}(\frac{13}{12})^2\times 4\pi=\frac{\frac{dx}{dt}12-3\times 5}{12^2}\end{equation}

    \begin{equation*}169\times4\pi=12\frac{dx}{dt}-15\end{equation}

    \begin{equation*}676\pi+15=12\frac{dx}{dt}\end{equation}

    \begin{equation*}\frac{dx}{dt}=178.2\end{equation}

The velocity of the light is 178.2 m/minute.

Leave a Comment

Filed under Differentiation, Implicit, Pythagoras, Right Trigonometry, Trigonometry, Uncategorized

Tagged as ,

October 1, 2024 · 7:31 am

Using De Moivre’s Theorem for Trigonometric Identities

We are going to use De Moivre’s theorem to prove trigonometric identities.

Remember, De Moivre’s Theorem

If z=r(cos(\theta)+isin(\theta)), then z^n=r^n(cos(n\theta)+isin(n\theta))

Or a shorter version z=rcis(\theta), then z^n=r^ncis(n\theta)

Now, let z=cos(\theta)+isin(\theta), find z+\frac{1}{z}

z+z^{-1}=cos(\theta)+isin(\theta)+cos(-\theta)+isin(-\theta)

Remember cos(\theta)=cos(\theta) and sin(-\theta)=-sin(\theta)

z+\frac{1}{z}=cos(\theta)+isin(\theta)+cos(\theta)-isin(\theta)

z+\frac{1}{z}=2cos(\theta)

It is the same for z^n+\frac{1}{z^n}

z^n+z^{-n}=cos(n\theta)+isin(n\theta)+cos(-n\theta)+isin(-n\theta)

z^n+\frac{1}{z^n}=2cos(n\theta)

Prove cos(2\theta)=2cos^2(\theta)-1
LHS=\frac{1}{2}(z^2+\frac{1}{z^2})
LHS=\frac{1}{2}(z^2+\frac{1}{z^2})+z\times\frac{1}{z}-z\times\frac{1}{z}
LHS=\frac{1}{2}(z^2+2z\times\frac{1}{z}+\frac{1}{z^2})-z\times\frac{1}{z}
LHS=\frac{1}{2}(z+\frac{1}{z})^2-1
LHS=\frac{1}{2}(2cos(\theta))^2-1
LHS=\frac{1}{2}(4cos^2(\theta))-1
LHS=2cos^2(\theta)-1
LHS=RHS

We can do something similar with sine.

z-\frac{1}{z}=cos(\theta)+isin(\theta)-(cos(-\theta)+isin(-\theta))

z-\frac{1}{z}=cos(\theta)+isin(\theta)-(cos(-\theta)+isin(-\theta))

z-\frac{1}{z}=cos(\theta)+isin(\theta)-(cos(\theta)-isin(\theta))

z-\frac{1}{z}=cos(\theta)+isin(\theta)-cos(\theta)+isin(\theta)

z-\frac{1}{z}=2isin(\theta)

Hence z^n-\frac{1}{z^n}=2isin(n\theta)

Prove sin(2\theta)=2sin(\theta)cos(\theta)
LHS=sin(2\theta)
LHS=\frac{1}{2i}(z^2-\frac{1}{z^2})
LHS=\frac{1}{2i}(z-\frac{1}{z})(z+\frac{1}{z})
LHS=\frac{1}{2i}(2isin(\theta)(2cos(\theta))
LHS=sin(\theta)2cos(\theta)
LHS=2sin(\theta)cos(\theta)
LHS=RHS

Let’s find an identity for cos(3\theta)

cos(3\theta)=\frac{1}{2}(z^3+\frac{1}{z^3})

=\frac{1}{2}(z^3+\frac{1}{z^3}+3z^2\times\frac{1}{z}+3z\times\frac{1}{z^2}-3z^2\times\frac{1}{z}-3z\times\frac{1}{z^2})

=\frac{1}{2}((z+\frac{1}{z})^3-3z-\frac{3}{z})

=\frac{1}{2}((z+\frac{1}{z})^3-3(z+\frac{1}{z}))

=\frac{1}{2}(2cos(\theta))^3-3(2cos(\theta)))

=\frac{1}{2}(8cos^3(\theta)-6cos(\theta))

=4cos^3(\theta)-3cos(\theta)

\therefore cos(3\theta)=4cos^3(\theta)-3cos(\theta)

And sin(3\theta)?

sin(3\theta)=\frac{1}{2i}(z^3-\frac{1}{z^3})

=\frac{1}{2i}(z^3-\frac{1}{z^3}-3z^2\times\frac{1}{z}+3z\times\frac{1}{z^2}+3z^2\times\frac{1}{z}-3z\times\frac{1}{z^2}

=\frac{1}{2i}((z-\frac{1}{z})^3+3z-\frac{3}{z})

=\frac{1}{2i}(2isin(\theta))^3+3(z-\frac{1}{z}))

=\frac{1}{2i}(-8isin^3(\theta)+6isin(\theta))

=-4sin^3(\theta)+3sin(\theta)

\therefore sin(3\theta)=3sin(\theta)-4sin^3(\theta)

Leave a Comment

Filed under Complex Numbers, Identities, Trig Identities, Trigonometry

Tagged as , ,

June 1, 2024 · 7:43 am

Non-Right Trigonometry Problem

I worked on this question with one of my students (I don’t know where it is from).


Mike leaves the rose bush he was examining and walks 35m in the direction
S20^{\circ}W towards a pond.
From there he walks 70m towards a rotunda. Mike is now 100m from the rose bush.
Find the bearing of the rotunda from the pond.

Let’s try to draw a diagram

Because we don’t the direction Mike walked from the pond, I have drawn a circle with radius 70m centred at the pond.

We know Mike is now 100m from the rose bush. As we don’t know the direction, I have drawn another circle with radius 100m centred at the rose bush. Where the two circles intersect are the possible locations of the rotunda.

First Position



Use the cosine rule to find the angle
cos\theta=\frac{b^2+c^2-a^2}{2bc}
cos\theta=\frac{70^2+35^2-100^2}{2(70)(35)}
\theta=cos^{-1}(\frac{-3875}{5250})
\theta=137.6^{\circ}




Using the fact that alternate angles in parallel lines are congruent, we can see
that the bearing from the pond to the rotunda is
360-(137.6-20)=242.2^{\circ} T

Second Position


It is the same triangle, so
\theta=137.6^{\circ}.

This time the bearing is
20+137.6=157.6^{\circ}T

Hence, the two possible bearings of the rotunda from the pond are 242.2^{\circ}T or 157.6^{\circ}T.

Leave a Comment

Filed under Bearings, Non-Right Trigonometry, Trigonometry

Tagged as ,

May 28, 2024 · 6:52 am

Proof of the Sine Rule

As I have done a cosine rule proof, I thought I should also do a sine rule proof.

\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}\textnormal{ or }\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}

From the above diagram, we can find h in two ways.

sinC=\frac{h}{a}

(1)   \begin{equation*}h=asinC\end{equation*}

sinA=\frac{h}{c}

(2)   \begin{equation*}h=csinA\end{equation*}

Set equation 1 equal to equation 2

asinC=csinA

\frac{a}{sinA}=\frac{c}{sinC} or \frac{sinC}{c}=\frac{sinA}{a}

We could have put the altitude of the triangle from vertex A

Following the same process as above

sinC=\frac{h}{b}

(3)   \begin{equation*}h=bsinC\end{equation*}

sinB=\frac{h}{c}

(4)   \begin{equation*}h=csinB\end{equation*}

Set equation 3 equal to equation 4.

bsinC=csinB

\frac{b}{sinB}=\frac{c}{sinC}

Now \frac{c}{sinC}=\frac{a}{sinA} therefore

\frac{b}{sinB}=\frac{c}{sinC}=\frac{a}{sinA} or \frac{sinB}{b}=\frac{sinC}{c}=\frac{sinA}{a}

Leave a Comment

Filed under Non-Right Trigonometry, Right Trigonometry, Trigonometry

Tagged as , ,

May 25, 2024 · 7:58 am

Proof of the cosine rule

    \[a^2=b^2+c^2-2bccos(A)\]

From the diagram above

    \[h^2=a^2-(b-x)^2\textnormal{ and }h^2=c^2-x^2\]

    \[\therefore a^2-(b-x)^2=c^2-x^2\]

    \[a^2-(b^2-2bx+x^2)=c^2-x^2\]

    \[a^2-b^2+2bx-x^2-c^2+x^2=0\]

(1)   \begin{equation*}a^2=b^2+c^2-2bx\end{equation*}

From the diagram above, we can see

    \[cos A=\frac{x}{c}\]

(2)   \begin{equation*}x=c cos A\end{equation*}

Substitute equation 2 into equation 1

    \[a^2=b^2+c^2-2bc cosA\]

It can also be handy to have the angle version

    \[cosA=\frac{b^2+c^2-a^2}{2bc}\]

1 Comment

Filed under Non-Right Trigonometry, Right Trigonometry, Trigonometry, Uncategorized

Tagged as ,

May 19, 2024 · 5:42 am

Right-Angled Trigonometry Question (very hard)

One of my year 10 students came with this question from his text book.

ICE_EM Mathematics Year 10 Third Edition

Here is my solution.

A pdf version of the solution

Leave a Comment

Filed under Right Trigonometry, Trigonometry, Uncategorized

Tagged as , ,

March 29, 2024 · 6:20 am

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.

Leave a Comment

Filed under Area, Area of Triangles (Sine), Heron's Law, Non-Right Trigonometry, Shoelace Forumla, Trigonometry

Tagged as , ,