Category Archives: Trigonometry

April 21, 2025 · 11:08 am

Area/Geometry Problem

This problem is from The Geometry Forum Problem of the Week June 1996

In triangle ABC, AC=18 and D is the point on AC for which AD=5. Perpendiculars drawn from D to AB and CB have lengths of 4 and 5 respectively. What is the area of triangle ABC?

I put together a diagram (in Geogebra)

Add points P and Q

Triangle APD and triangle DQC are right angled. Using pythagoras, $AP=3$ and $QC=12$

$BQDP$ is a cyclic quadrilateral and $BD$ is the diameter. I am not sure if this is useful, but it is good to notice.

$\begin{equation*}sin(A+B+C)=sin(180)=0\end{equation}$

$\begin{equation*}sin((A+C)+B)=sin(A+C)cosB+sinBcos(A+C)=0\end{equation}$

$\begin{equation*}cosB(sinAcosC+sinCcosA)+sinB(cosAcosC-sinAsinC)=0\end{equation}$

$\begin{equation*}cosB(\frac{4}{5}\times\frac{12}{13}+\frac{5}{13}\times\frac{3}{5})+sinB(\frac{3}{5}\times\frac{12}{13}-\frac{4}{5}\times\frac{5}{13})=0\end{equation}$

$\begin{equation*}cosB(\frac{48}{65}+\frac{15}{65})+sinB(\frac{36}{65}-\frac{20}{65})=0\end{equation}$

$\begin{equation*}\frac{63}{65}cosB+\frac{16}{65}sinB=0\end{equation}$

$\begin{equation*}63cosB+16sinB=0\end{equation}$

$\begin{equation*}63+16tanB=0\end{equation}$

$\begin{equation*}tanB=\frac{-63}{16}\end{equation}$

If $tanB=\frac{-63}{16}$ then $sinB=\frac{63}{65}$

Now,

$\begin{equation*}\frac{y+12}{sinA}=\frac{18}{sinB}\end{equation}$

$\begin{equation*}y+12=\frac{4}{5}(18)\frac{65}{63}\end{equation}$

$\begin{equation*}y+12=\frac{104}{7}\end{equation}$

Hence the Area is

$\begin{equation*}A=\frac{1}{2}(18)(\frac{104}{7})sinC\end{equation}$

$\begin{equation*}A=\frac{1}{2}(18)(\frac{104}{7})\frac{5}{13}\end{equation}$

$\begin{equation*}A=\frac{360}{7}=51.43\end{equation}$

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Filed under Area, Finding an area, Geometry, Identities, Non-Right Trigonometry, Pythagoras, Trigonometry

Tagged as area, area of triangles, problem solving

April 7, 2025 · 9:09 am

The vector projection (vector resolution or vector component) of $\mathbf{a}$ onto a non-zero vector $\mathbf{b}$ is splitting $\mathbf{a}$ into two vectors, one is parallel to $\mathbf{b}$ (the vector projection) and one perpendicular to $\mathbf{b}$

In the above diagram $\mathbf{a_1}$ is the vector projection of $\mathbf{a}$ onto $\mathbf{b}$ and $\mathbf{a_2}$ is perpendicular to $\mathbf{b}$ .

How do we find $\mathbf{a_1}$ and $\mathbf{a_2}$ ?

Using right trigonometry,

$cos(\theta)=\frac{|\mathbf{a_1}|}{|\mathbf{a}|}$

Remember the scalar product (dot product) of vectors is

(1) $\begin{equation*}\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|cos(\theta)\end{equation*}$

Hence $cos(\theta)=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$

$\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{|\mathbf{a_1}|}{|\mathbf{a}|}$

and, $|\mathbf{a_1}|=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\times|\mathbf{a}|$

$|\mathbf{a_1}|=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$

This is the scalar projection of $\mathbf{a}$ onto $\mathbf{b}$

To find the vector projection we need to multiply by $\mathbf{\hat{b}}$ , that is find a vector with the same magnitude as $\mathbf{a_1}$ in the direction of $\mathbf{b}$ .

The vector projection is

$\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\times \mathbf{\hat{b}}$

$\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\times \frac{\mathbf{b}}{|\mathbf{b}|}$

$\mathbf{a_1}=\frac{\mathbf{a}\cdot\mathbf{b}}{(|\mathbf{b}|)^2}\times \mathbf{b}$

Now for $\mathbf{a_2}$ , we know $\mathbf{a}=\mathbf{a_1}+\mathbf{a_2}$

Hence, $\mathbf{a_2}=\mathbf{a}-\mathbf{a_1}$

Example

From Cambridge Year 11 Specialist Mathematics (Chapter 3)

(a) $\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\mathbf{b}$

$=\frac{\begin{pmatrix}4\\1\end{pmatrix}\cdot\begin{pmatrix}1\\-1\end{pmatrix}}{2}\times(\mathbf{i}-\mathbf{j})$

$=\frac{3}{2}{(\mathbf{i}-\mathbf{j})$

(b) $4\mathbf{i}+\mathbf{j}-\frac{3}{2}{(\mathbf{i}-\mathbf{j})=\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}$

(c)

The shortest distance (green vector) is the vector component of $\mathbf{a}$ perpendicular to $\mathbf{b}$ , i.e. $\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}$

$|\frac{5}{2}\mathbf{i}+\frac{5}{2}\mathbf{j}|=\sqrt{\frac{25}{4}+\frac{25}{4}}=\sqrt{\frac{50}{4}}=\frac{5\sqrt{2}}{2}$

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Filed under Right Trigonometry, Trigonometry, Vector Projection, Vectors, Year 11 Specialist Mathematics

Tagged as dot product, scalar product, scalar projection, vector projection

April 3, 2025 · 7:28 am

Area of Regular Polygons

Finding the area of a regular polygon when you know the side length

Find the area of an $n-$ sided regular polygon if you know the side length, $l$ .

Find the $h$ of the triangle in terms of $l$ and theta.

$tan(\theta)=\frac{\frac{l}{2}}{h}$

$h=\frac{l}{2tan(\theta)}$

Remember the area of a triangle is $A=\frac{1}{2}bh$

Hence, $A=\frac{1}{2} l \times \frac{l}{2tan(\theta)}=\frac{l^2}{4tan(\theta)}$

And $\theta=\frac{360}{2n}=\frac{180}{n}$

Therefore $A=\frac{l^2}{4tan(\frac{180}{n})}$

There are $n$ triangles in an $n-$ sided polygon

(1) $\begin{equation*}A=\frac{nl^2}{4tan(\frac{180}{n})}\end{equation*}$

A=\frac{6\times10^2}{4tan(\frac{180}{6})}

Finding the area of a polygon if you know the inradius or the apothem

The apothem and the inradius are the same. It is the radius of the incircle.

Find the area of the triangle in terms of $a$ and theta.

$tan(\theta)=\frac{\frac{l}{2}}{a}$

$l=2atan(\theta)$

$A=\frac{1}{2}bh$

$A=\frac{1}{2}2atan(\theta)a=a^2tan(\theta)$

And $\theta=\frac{180}{n}$

Hence for an $n-$ sided polygon

(2) $\begin{equation*}A=na^2tan(\frac{180}{n})\end{equation*}$

Finding the area of a regular polygon given the circumradius

The circumradius is the radius of the circumscribed circle ( $R$ in the diagram above)

Remember the area of triangle formula

$A=\frac{1}{2}absin(\theta)$

$A=\frac{1}{2}R^2sin(\theta)$

$\theta=\frac{360}{n}$

Hence, $A=\frac{1}{2}R^2sin(\frac{360}{n})$

Hence, for an $n-$ sided polygon

(3) $\begin{equation*}A=\frac{nR^2sin(\frac{360}{n})}{2}\end{equation*}$

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Filed under Area, Area of Triangles (Sine), Finding an area, Non-Right Trigonometry, Regular Polygons, Right Trigonometry, Year 11 Mathematical Methods

Tagged as area of regular polygons, area of triangles, trig

February 4, 2025 · 9:17 am

Differentiating the Tangent Function

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

I use the quotient rule to differentiate $f(x)=tan(x)$ .

(1) $\begin{equation*}\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}\end{equation*}$

If $h(x)=tan(x)=\frac{sin(x)}{cos(x)}$ then from equation $1$

(2) $\begin{equation*}h'(x)=\frac{cos(x)cos(x)-(-sin(x)sin(x))}{[cos(x)]^2}\end{equation*}$

(3) $\begin{equation*}h'(x)=\frac{cos^2(x)+sin^2(x)}{cos^2(x)}\end{equation*}$

Remember the Pythagorean identity

(4) $\begin{equation*}sin^2(x)+cos^2(x)=1\end{equation*}$

Hence

$\begin{equation*}h'(x)=\frac{1}{cos^2(x)}=sec^2(x)\end{equation}$

(5) $\begin{equation*}\frac{d}{dx}tan(x)=sec^2(x)\end{equation*}$

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Filed under Calculus, Differentiation, Differentiation, Identities, Quotient Rule, Trigonometry, Year 12 Mathematical Methods

Tagged as differentiating tan, differentiating trig

February 3, 2025 · 9:11 am

Differentiating Trigonometric Functions

In the last post we looked at two trig limits:

(1) $\begin{equation*}\lim_{x \to 0}\frac{sin(x)}{x}=1\end{equation*}$

(2) $\begin{equation*}\lim_{x \to 0}\frac{1-cos(x)}{x}=0\end{equation*}$

We are going to use these two limits to differentiate sine and cosine functions from first principals.

$\begin{equation*}f(x)=sin(x)\end{equation}$

$\begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{sin(x+h)-sin(x)}{h}\end{equation}$

Use the trig identity

$\begin{equation*}sin(A+B)=sinAcosB+sinBcosA\end{equation}$

$\begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{sin(x)cos(h)+sin(h)cos(x)-sin(x)}{h}\end{equation}$

$\begin{equation*}f'(x)=\lim\limits_{h \to 0}(\frac{sin(x)(cos(h)-1)}{h}+\frac{sin(h)cos(x)}{h})\end{equation}$

$\begin{equation*}f'(x)=sin(x)\lim\limits_{h \to 0}(\frac{(cos(h)-1)}{h}+cos(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}$

$\begin{equation*}f'(x)=sin(x)\lim\limits_{h \to 0}(\frac{-(-cos(h)+1)}{h}+cos(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}$

Evaluate the limits

$\begin{equation*}f'(x)=sin(x)\times 0+cos(x)\times (1)=cos(x)\end{equation}$

Hence, $\frac{d}{dx}sin(x)=cos(x)$ .

Now we are going to do the same for $f(x)=cos(x)$ .

$\begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x+h)-cos(x)}{h}\end{equation}$

Use the trigonometric identity

$\begin{equation*}cos(A+B)=cosAcosB-sinAsinB\end{equation}$

$\begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x)cos(h)-sin(x)sin(h)-cos(x)}{h}\end{equation}$

$\begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x)(cos(h)-1)-sin(x)sin(h)}{h}\end{equation}$

$\begin{equation*}f'(x)=cos(x)\lim\limits_{h \to 0}\frac{-(1-cos(h))}{h}-sin(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}$

Evaluate the limits

$\begin{equation*}f'(x)=cos(x)\times(0)-sin(x)\times (1)=-sin(x)\end{equation}$

Hence $\frac{d}{dx} cos(x)=-sin(x)$

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Filed under Calculus, Differentiation, Identities, Trigonometry, Year 12 Mathematical Methods

Tagged as differentiation, trig differentiation, trig identities, trig limits, trigonometric differentiation, trigonometric identities, trigonometric limits

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 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 $tan(\beta)=n-1$ .

tan(\alpha-\beta)=\frac{tan(\alpha)-tan(\beta)}{1+tan(\alpha)tan(\beta)}

tan(\alpha-\beta)=\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}

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.

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

Tagged as limits, sum, trigonometric identities

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

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

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

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

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

Tagged as infinite product expansion of cos(x), trigonometry

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 $z^n+\frac{1}{z^n}$

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

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

We can do something similar with sine.

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

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

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

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

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

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 $sin(3\theta)$ ?

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

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

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

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

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

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

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

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Filed under Complex Numbers, Identities, Trig Identities, Trigonometry

Tagged as complex numbers, de moivre, trig identities

Category Archives: Trigonometry

Area/Geometry Problem

Vector and Scalar Projection

Example

Area of Regular Polygons

Finding the area of a regular polygon when you know the side length

Finding the area of a polygon if you know the inradius or the apothem

Finding the area of a regular polygon given the circumradius

Differentiating the Tangent Function

Differentiating Trigonometric Functions

Trigonometric Limits

Interesting Sum

Infinite Product Expansion of cos (x)

Using De Moivre’s Theorem for Trigonometric Identities

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