Category Archives: Trigonometry

October 9, 2025 · 6:37 am

Trigonometric Exact Values

Find exactly $sin(18^\circ)$

We must be able to find an arithmetic combination of the exact values we knew to find $18$ .

$\begin{equation*}90=5\times 18\end{equation}$

$\begin{equation*}90-3(18)=2(18)\end{equation}$

I re-arranged as above, so I could take advantage of $cos(90)=0$ and $sin(90)=1$

cos(2x)=cos^2(x)-sin^2(x)=1-2sin^2(x)=2cos^2(x)-1

$\begin{equation*}sin(90-3(18))=sin(2(18))\end{equation}$

$\begin{equation*}sin(90)cos(3(18))-sin(3(18))cos(90)=2sin(18)cos(18)\end{equation}$

$\begin{equation*}cos(3(18))=2sin(18)cos(18)\end{equation}$

$\begin{equation*}4cos^3(18)-3cos(18)=2sin(18)cos(18)\end{equation}$

$\begin{equation*}4cos^3(18)-3cos(18)-2sin(18)cos(18)=0\end{equation}$

$\begin{equation*}cos(18)(4cos^2(18)-3-2sin(18))=0\end{equation}$

Hence,

$\begin{equation*}4cos^2(18)-2sin(18)-3=0\end{equation}$

$\begin{equation*}4-4sin^2(18)-2sin(18)-3=0\end{equation}$

$\begin{equation*}-4sin^2(18)-2sin(18)+1=0\end{equation}$

Use the quadratic equation formula

$\begin{equation*}sin(18)=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{equation}$

$\begin{equation*}sin(18)=\frac{2 \pm \sqrt{4-4(-4)(1)}}{-8}\end{equation}$

$\begin{equation*}sin(18)=\frac{2 \pm \sqrt{20}}{-8}\end{equation}$

$\begin{equation*}sin(18)=\frac{-2 \mp 2\sqrt{5}}{8}\end{equation}$

$\begin{equation*}sin(18)=\frac{-1 \mp \sqrt{5}}{4}\end{equation}$

As $sin(18)>0$ , $sin(18)=\frac{-1+\sqrt{5}}{4}$

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Filed under Addition and Subtraction Identities, Algebra, Identities, Quadratic, Quadratics, Solving, Solving Equations, Solving Trig Equations, Trigonometry, Year 11 Specialist Mathematics

Tagged as exact values, sin(18), trig exact values, trigonometric identities

September 30, 2025 · 3:54 am

Geometry Problem

*Geometry Snacks* by Ed Southall and Vincent Pantaloni

I started by trisecting another side of the triangle

This makes it clearer that the two lines are parallel

Which means the two angles labelled above are corresponding and therefore congruent.

Let the side length be $x$ .

The area of the equilateral triangle is

(1) $\begin{equation*}A=\frac{1}{2}x^2 sin(60)=\frac{\sqrt{3}x^2}{4}\end{equation*}$

$\begin{equation*}cos(60)=\frac{y}{\frac{2x}{3}}\end{equation}$

$\begin{equation*}y=\frac{x}{3}\end{equation}$

Area of right triangle

(2) $\begin{equation*}A=(\frac{1}{2})(\frac{2x}{3})(\frac{x}{3})sin(60)=\frac{\sqrt{3}x^2}{18}\end{equation*}$

The fraction of the area is

$\begin{equation*}=\frac{\frac{\sqrt{3}x^2}{18}}{\frac{\sqrt{3}x^2}{4}}=\frac{2}{9}\end{equation}$

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Filed under Algebra, Area, Area of Triangles (Sine), Finding an area, Geometry, Puzzles, Right Trigonometry, Simplifying fractions, Trigonometry

Tagged as geometry problem, geometry snacks

August 25, 2025 · 3:29 am

Matrices -Linear Transformations (Rotation)

We are going to find a matrix to rotate a point about the origin a number of degrees (or radians).

$P$ and $P'$ are equidistant from the origin. I.e. $\sqrt{c^2+d^2}=\sqrt{a^2+b^2}$

Remember, anti-clockwise angles are positive.

$\begin{equation*}cos(\theta+\alpha)=\frac{c}{\sqrt{c^2+d^2}}\end{equation}$

$\begin{equation*}c=\sqrt{a^2+b^2}cos(\theta+\alpha)\end{equation}$

Use the cosine addition identity.

$\begin{equation*}c=\sqrt{a^2+b^2}(cos(\theta)cos(\alpha)-sin(\theta)sin(\alpha))\end{equation}$

$\begin{equation*}c=\sqrt{a^2+b^2}(cos(\theta)\frac{a}{\sqrt{a^2+b^2}}-sin(\theta)\frac{b}{\sqrt{a^2+b^2}})\end{equation}$

(1) $\begin{equation*}c=acos(\theta)-bsin(\theta)\end{equation*}$

We will do the same for $d$

$\begin{equation*}sin(\theta+\alpha)=\frac{d}{\sqrt{a^2+b^2}}\end{equation}$

$\begin{equation*}d=\sqrt{a^2+b^2}sin(\theta+\alpha)\end{equation}$

Use the sine addition identity.

$\begin{equation*}d=\sqrt{a^2+b^2}(sin(\theta)cos(\alpha)+cos(\theta)sin(\alpha)\end{equation}$

$\begin{equation*}d=\sqrt{a^2+b^2}(sin(\theta)\frac{a}{\sqrt{a^2+b^2}}+cos(\theta)\frac{b}{\sqrt{a^2+b^2}})\end{equation}$

(2) $\begin{equation*}d=asin(\theta)+bcos(\theta)\end{equation*}$

Let $R$ be the rotation matrix, then

$\begin{equation*}R\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}acos(\theta)-bsin(\theta)\\asin(\theta)+bcos(\theta)\end{bmatrix}\end{equation}$

Hence $R$ must be

(3) $\begin{equation*}R=\begin{bmatrix}cos(\theta)&-sin(\theta)\\sin(\theta)&cos(\theta)\end{bmatrix}\end{equation*}$

Example

Find the image of the line $y=x+1$ after it is rotated $60^\circ$ about the origin.

I am going to select two points on the line and transform them.

$\begin{equation*}\begin{bmatrix}cos(60)&-sin(60)\\sin(60)&cos(60)\end{bmatrix}\begin{bmatrix}0&4\\1&5\end{bmatrix}=\begin{bmatrix}x'_1&x'_2\\y'_1&y'_2\end{bmatrix}\end{equation}$

$\begin{equation*}\begin{bmatrix}\frac{1}{2}&-\frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2}&\frac{1}{2}\end{bmatrix}\begin{bmatrix}0&4\\1&5\end{bmatrix}=\begin{bmatrix}x'_1&x'_2\\y'_1&y'_2\end{bmatrix}\end{equation}$

$\begin{equation*}\begin{bmatrix}-\frac{\sqrt{3}}{2}&\frac{4-5\sqrt{3}}{2}\\\frac{1}{2}&2\sqrt{3}+\frac{5}{2}\end{bmatrix}=\begin{bmatrix}x'_1&x'_2 \\y'_1&y'_2\end{bmatrix}\end{equation}$

We can then find the equation of the line.

$\begin{equation*}m=\frac{2\sqrt{3}+\frac{5}{2}-\frac{1}{2}}{\frac{4-5\sqrt{3}+\sqrt{3}}{2}}\end{equation}$

$\begin{equation*}m=-2-\sqrt{3}\end{equation}$

$\begin{equation*}y-\frac{1}{2}=(-2-\sqrt{3})(x+\frac{\sqrt{3}}{2})\end{equation}$

$\begin{equation*}y=(-2-\sqrt{3})x-1-\sqrt{3}\end{equation}$

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Filed under Addition and Subtraction Identities, Identities, Matrices, Transformations, Trigonometry, Year 11 Specialist Mathematics

Tagged as matrix transformations, rotation matrix

August 14, 2025 · 6:39 am

Trig Identities and Exact Values

My Year 11 Specialist Mathematics students are working on Trig identities. We came across this question

Without the use of a calculator, evaluate
(a) $cos20^\circ\times cos40^\circ\times cos80^\circ$

(b) $cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})$

OT Lee Year 11 Specialist Mathematics textbook

I spent a bit of time thinking about the question. Can you use a product to sum identity twice? But I was always being left with an angle that doesn’t have a nice exact value.

I tried a few things, had a chat to Meta AI, and finally stumbled upon this method.

Remember

$\begin{equation*}sin(2x)=2sin(x)cos(x)\end{equation}$

Which can be rearranged to

$\begin{equation*}cos(x)=\frac{sin(2x)}{sin(x)}\end{equation}$

(a) $cos20^\circ\times cos40^\circ\times cos80^\circ=\frac{sin(40)}{2sin(20)}\frac{sin(80)}{2sin(40)}\frac{sin(160)}{2sin(80)}$

Which simplifies to

$\begin{equation*}\frac{sin(160)}{8sin(20)}\end{equation}$

Now $sin(160)=sin(20)$

Hence $cos20^\circ\times cos40^\circ\times cos80^\circ=\frac{1}{8}$

And we will do the same for part (b)

$cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})=\frac{sin(\frac{2\pi}{7})}{2sin(\frac{\pi}{7})}\frac{sin(\frac{4\pi}{7})}{2sin(\frac{2\pi}{7})}\frac{sin(\frac{8\pi}{7})}{2sin(\frac{4\pi}{7})}$

Which simplifies to

$cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})=\frac{sin(\frac{8\pi}{7})}{8sin(\frac{\pi}{7})}$

Now $sin(\frac{8\pi}{7})=-sin(\frac{\pi}{7})$

Hence $cos(\frac{\pi}{7})\times cos(\frac{2\pi}{7})\times cos(\frac{4\pi}{7})=\frac{-1}{8}$

And then I had to test them on my Classpad.

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Filed under Classpad Skills, Identities, Simplifying fractions, Trigonometry, Year 11 Specialist Mathematics

Tagged as exact values, trig identities

August 10, 2025 · 4:54 am

Trigonometric Identities – Product to Sum

Let’s think about the sine and cosine addition and subtraction trig identities.

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

(2) $\begin{equation*}sin(A-B)=sinAcosB-cosAsinB\end{equation*}$

If we add equation $1$ and $2$ , we get

$\begin{equation*}sin(A+B)+sin(A-B)=2sinAcosB\end{equation}$

Hence, $sinAcosB=\frac{1}{2}(sin(A+B)+sin(A-B))$

If we subtract equation $2$ from equation $1$ , we get

$\begin{equation*}sin(A+B)-sin(A-B)=2cosAsinB\end{equation}$

Hence, $cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B)$

What about the cosine addition and subtraction idenities?

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

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

If we add equation $3$ and $4$ , we get

$\begin{equation*}cos(A+B)+cos(A-B)=2cosAcosB\end{equation}$

Hence, $cosAcosB=\frac{1}{2}(cos(A+B)+cos(A-B))$

If we subtract $3$ from $4$ , we get

$\begin{equation*}cos(A-B)-cos(A+B)=2sinAsinB\end{equation}$

Hence, $sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))$

These are the product to sum identities.

$\begin{equation*}sinAcosB=\frac{1}{2}(sin(A+B)+sin(A-B))\end{equation}$

$\begin{equation*}cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B))\end{equation}$

$\begin{equation*}cosAcosB=\frac{1}{2}(cos(A+B)+cos(A-B))\end{equation}$

$\begin{equation*}sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))\end{equation}$

Examples

(1) Solve $sin(5x)-sin(x)=0$ for $0\le x \le 2\pi$

Remember,

$\begin{equation*}cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B))\end{equation}$

$\begin{equation*}A+B=5\end{equation}$

$\begin{equation*}A-B=1\end{equation}$

Therefore, $A=3$ and $B=2$

$\begin{equation*}sin(5x)-sin(x)=2cos(3x)sin(2x)=0\end{equation}$

$\begin{equation*}cos(3x)=0\end{equation}$

$\begin{equation*}3x=\frac{\pi}{2},\frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{\9\pi}{2}, \frac{11\pi}{2}\end{equation}$

$\begin{equation*}x=\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}\end{equation}$

$\begin{equation*}sin(2x)=0\end{equation}$

$\begin{equation*}2x=0, \pi, 2\pi, 3\pi, 4\pi\end{equation}$

$\begin{equation*}x=0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\end{equation}$

Hence $x=0, \frac{\pi}{6}. \frac{\pi}{2}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}, 2\pi$

(2)Solve $sin(7\theta)-sin(\theta)=sin(3\theta)$ for $0 \le \theta \le\2\pi$

$\begin{equation*}cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B))\end{equation}$

Therefore, $A+B=7$ and $A-B=1$

$A=4, B=3$

$\begin{equation*}2cos(4\theta)sin(3\theta)=sin(3\theta)\end{equation}$

$\begin{equation*}2cos(4\theta)sin(3\theta)-sin(3\theta)=0\end{equation}$

$\begin{equation*}sin(3\theta)(2cos(4\theta)-1)=0\end{equation}$

$sin(3\theta)=0$ and $cos(4\theta)=\frac{1}{2}$

$3\theta=0, \pi, 2\pi, 3\pi, 4\pi, 5\pi, 6\pi$

$\theta=0, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}. 2\pi$

$cos(4\theta)=\frac{1}{2}$

$4\theta=\frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}, \frac{13\pi}{3}, \frac{17\pi}{3}, \frac{19\pi}{3}, \frac{23\pi}{3}$

$\theta=\frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}, \frac{23\pi}{12}$

Hence $\theta=0, \frac{\pi}{3}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{5\pi}{3}, \frac{23\pi}{12}, 2\pi$

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Filed under Addition and Subtraction Identities, Identities, Product to Sum idenitites, Trigonometry

Tagged as product to sum trig identities, solving trig equations, trig identities

August 2, 2025 · 9:30 am

Trigonometric Exact Value

Using an appropriate double angle identity, find the exact value of
$cos(\frac{\pi}{12})$

The double angle identity for sine is

(1) $\begin{equation*}cos(2A)=cos^2A-sin^2A=2cos^2A-1=1-2sin^2A\end{equation*}$

That means $\frac{\pi}{12}$ is either $2A$ or $A$ .

It must be $A$ as $2\times\frac{\pi}{12}=\frac{\pi}{6}$ as there are exact values for $\frac{\pi}{6}$

Hence,

$\begin{equation*}cos{\frac{\pi}{6}}=2cos^2{\frac{\pi}{12}}-1\end{equation}$

$\begin{equation*}\frac{\sqrt{3}}{2}=2cos^2{\frac{\pi}{12}}-1\end{equation}$

$\begin{equation*}\frac{\sqrt{3}}{2}+1=2cos^2{\frac{\pi}{12}}\end{equation}$

$\begin{equation*}\frac{\frac{\sqrt{3}}{2}+1}{2}=cos^2{\frac{\pi}{12}}\end{equation}$

$\begin{equation*}\frac{\sqrt{3}+2}{4}=cos^2{\frac{\pi}{12}}\end{equation}$

$\begin{equation*}\sqrt{\frac{\sqrt{3}+2}{4}}=cos{\frac{\pi}{12}}\end{equation}$

As $\frac{\pi}{12}$ is in the first quadrant, we don’t need to consider the negative version.

$\begin{equation*}cos(\frac{\pi}{12})=\frac{\sqrt{3}+2}{2}\end{equation}$

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Filed under Algebra, Identities, Trigonometry, Year 11 Mathematical Methods

Tagged as trig double angle identities, trig exact values

July 28, 2025 · 8:09 am

Trig Identities – Addition and Subtraction

Deriving the addition and subtraction trigonometric identities.

We will start with cosine, and use the result to derive the remaining identities.

Proving $cos(A-B)=cos(A)cos(B)+sin(A)(sin(B)$ .

$A$ and $B$ are represented in the unit circle below.

Remember $P(x_1,y_1)=(cos(A), sin(A))$ and $Q(x_2, y_2)=(cos(B), sin(B))$

Using the cosine rule and triangle $OPQ$ , find $PQ$

$\begin{equation*}(PQ)^2=1^2+1^2-2(1)(1)cos(A-B)\end{equation}$

$\begin{equation*}(PQ)^2=2-2cos(A-B)\end{equation}$

Using the distance between points, find $PQ$

$\begin{equation*}(PQ)^2=(x_1-x_2)^2+(y_1-y_2)^2\end{equation}$

$\begin{equation*}(PQ)^2=(cosA-cosB)^2+(sinA-sinB)^2\end{equation}$

$(PQ)^2=cos^2A-2cosAcosB+cos^2B+sin^2A-2sinAsinB+sin^2B$

Remember the Pythagorean identity

$\begin{equation*}cos^2\theta+sin^2\theta=1\end{equation}$

$\begin{equation*}(PQ)^2=2-2cosAcosB-2sinAsinB\end{equation}$

Hence

$\begin{equation*}2-2cos(A-B)=2-2cosAcosB-2sinAsinB\end{equation}$

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

We can then use this identity to find $cos(A+B)$ .

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

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

Remember $cos(-B)=cos(B)$ and $sin(-B)=-sin(B)$

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

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

We can also find $sin(A+B)$

Remember, $sin\theta=cos(\frac{\pi}{2}-\theta)$

$\begin{equation*}sin(A+B)=cos(\frac{\pi}{2}-(A+B))\end{equation}$

$\begin{equation*}sin(A+B)=cos((\frac{\pi}{2}-A)-B)\end{equation}$

$\begin{equation*}sin(A+B)=cos(\frac{\pi}{2}-A)cosB+sin(\frac{\pi}{2}-A)sinB\end{equation}$

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

We can use equation $2$ to find $sin(A-B)$

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

$\begin{equation*}sin(A+(-B))=sinAcos(-B)+cosAsin(-B)\end{equation}$

$\begin{equation*}sin(A+(-B))=sinAcos(B)-cosAsin(B)\end{equation}$

(4) $\begin{equation*}sin(A-B)=sinAcos(B)-cosAsin(B)\end{equation*}$

And we can use both the sine and cosine identities to find $tan(A+B)$

Remember $tan\theta=\frac{sin\theta}{cos\theta}$

$\begin{equation*}tan(A+B)=\frac{sin(A+B)}{cos(A+B)}\end{equation}$

$\begin{equation*}=\frac{sinAcosB+cosAsinB}{cosAcosB-sinAsinB}\end{equation}$

$\begin{equation*}=\frac{sinAcosB+cosAsinB}{cosAcosB-sinAsinB}\times \frac{cosAcosB}{cosAcosB}\end{equation}$

$\begin{equation*}=\frac{\frac{sinAcosB}{cosAcosB}+\frac{cosAsinB}{cosAcosB}}{\frac{cosAcosB}{cosAcosB}+\frac{sinAsinB}{cosAcosB}}\end{equation}$

$\begin{equation*}=\frac{tanA+tanB}{1-tanAtanB}\end{equation}$

(5) $\begin{equation*}tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}\end{equation*}$

and

(6) $\begin{equation*}tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}\end{equation*}$

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

$\begin{equation*}sin(A\pm B)=sinAcosB\pm cosAsinB\end{equation}$

$\begin{equation*}tan(A \pm B)=\frac{tan A \pm tanB}{1 \mp tanAtanB}\end{equation}$

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Filed under Addition and Subtraction Identities, Identities, Non-Right Trigonometry, Simplifying fractions, Trigonometry, Year 11 Mathematical Methods, Year 11 Specialist Mathematics

Tagged as addition and subtraction trig identities, trig identities

July 26, 2025 · 8:45 am

General Solutions to Trigonometric Equations

Solve $sinx=\frac{1}{2}$ for $0\le x \le 2\pi$

Sine is positive in the first and second quadrants.

$\begin{equation*}sinx=\frac{1}{2}\end{equation}$

$\begin{equation*}x=\frac{\pi}{6} \text{ and } x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\end{equation}$

But what if we aren’t given a domain for the $x$ values?

Then we need to give general solutions.

For example,

Solve $sinx=\frac{1}{2}$

As you can see from the sketch above, there are infinite solutions.

The sine function has a period of $360^\circ$ , and so if $\frac{\pi}{6}$ is a solution then $\2pi+\frac{\pi}{6}$ is also a solution. This means $\frac{\pi}{6}+2\pi n, n\in\mathbb{Z}$ is a general solution. And we can do the same for the second solution $\frac{5\pi}{6}+2\pi n$ .

In general

$\begin{equation*}sinx=y\end{equation}$

$\begin{equation*}x=arcsin(y)+2\pi n \text { and } x=\pi-arcsin(y)+2\pi n \end{equation}$

$\begin{equation*}x=arcsin(y)+2\pi n \text { and } x=\pi(2n+1)-arcsin(y), n \in \mathbb{Z}\end{equation}$

We can turn this into one equation
$\begin{equation*}x=(-1)^n arcsin(y)+n\pi, n \in \mathbb{Z}\end{equation}$

What about cosine?

Solve $cosx=\frac{1}{2}$

Cosine is positive in the first and fourth quadrants (it also has a period of $2\pi$ . The first two (positive) solutions are $\frac{\pi}{3}$ and $2\pi-\frac{\pi}{3}$ .

To generalise, $x=2\pi n+\frac{\pi}{3} \text { and }x=2\pi n -\frac{\pi}{3}$ , which we can make into one equation $x=2\pi n \pm \frac{pi}{3}$

In general

$\begin{equation*}cosx=y\end{equation}$

$\begin{equation*}x=2\pi n \pm arccos(y), n\in\mathbb{Z}\end{equation}$

What about the tangent function? Remember tan has a period of $\pi$ .

Solve $tanx=\sqrt{3}$

First, note that the solutions are all a common distance ( $\pi$ ) apart.

Tan is positive in the first and the third quadrant

$\begin{equation*}tanx=\sqrt{3}\end{equation}$

$\begin{equation*}x=\frac{\pi}{3} \text { and } x=\pi+\frac{\pi}{3}\end{equation}$

Because all of the solutions are $\pi$ radians apart, the general solution is $x=\frac{\pi}{3} \pm \pi$

In general

$\begin{equation*}tanx=y\end{equation}$

$\begin{equation*}x=arctan(y) + n\pi, n\in \mathbb{Z}\end{equation}$

Examples

Solve for all values of $x$ , $tan^2(x)+tan(x)-6=0$

$\begin{equation*}tan^2(x)+tan(x)-6=0\end{equation}$

This is a quadratic equation – we need two numbers that add to $1$ and multiple to $-6$ , $+3 \text { and } -2$

$\begin{equation*}(tan(x)+3)(tan(x)-2))=0\end{equation}$

$\begin{equation*}tan(x)=-3 \text { or } tan(x)=2\end{equation}$

$\begin{equation*}x=arctan(-3)+n\pi \text { or } x=arctan(2)+n\pi, n\in\mathbb{Z}\end{equation}$

Solve $2cos(2x+\frac{\pi}{18})=\sqrt{3}$

$\begin{equation*}2cos(2x+\frac{\pi}{18})=\sqrt{3}\end{equation}$

$\begin{equation*}cos(2x+\frac{\pi}{18})=\frac{\sqrt{3}}{2}\end{equation}$

$\begin{equation*}2x+\frac{\pi}{18}=2n\pi \pm \frac{\pi}{6}\end{equation}$

$\begin{equation*}2x=2n\pi \pm \frac{\pi}{6}-\frac{\pi}{18}\end{equation}$

$\begin{equation*}2x=2n\pi \pm \frac{\pi}{9}\end{equation}$

$\begin{equation*}x=n\pi \pm \frac{\pi}{18}\end{equation}$

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Filed under Algebra, Quadratic, Solving Equations, Solving Trig Equations, Trigonometry, Year 11 Specialist Mathematics

Tagged as general solution to trig equations, solving trig equations

June 26, 2025 · 7:22 am

Trigonometry Question

I don’t know where I found this question, but it does require algebra and problem solving (as well as right trig and Pythagoras)

From a point $A$ , a lighthouse is on a bearing of $026^\circ$ T and the top of the light house is at an angle of elevation of $20.25^\circ$ . From a point $B$ , the lighthouse is on a bearing of $296^\cric$ T and the top of the lighthouse is at angle of elevation of $10.2^\circ$ . If $A$ and $B$ are 500 metres apart, find the height of the lighthouse.

Let’s draw a diagram.

Let the height of the lighthouse be $h$

We can find the angle between $A$ , the lighthouse, and $B$ by using the base triangle

The red line from $L$ is parallel to the two north lines. Hence $\theta=26^\circ+64^\circ=90^\circ$ (Alternate angles in parallel lines are congruent)

It’s a right triangle so we know

(1) $\begin{equation*}500^2=(AL)^2+(BL)^2\end{equation*}$

We are going to use the other two triangles to find $AL$ and $BL$

Substitute $AL$ and $BL$ into equation $1$

$\begin{equation*}500^2=(\frac{h}{tan(20.25)})^2+(\frac{h}{tan(10.2)})^2\end{equation}$

Solve for $h$ .

$\begin{equation*}500^2=7.35h^2+30.89h^2=38.24h^2\end{equation}$

$\begin{equation*}h^2=6538.3\end{equation}$

$\begin{equation*}h=80.9m\end{equation}$

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Filed under Algebra, Bearings, Pythagoras, Right Trigonometry, Solving Equations, Trigonometry

Tagged as hard trig, trigonometry question

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.

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

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

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

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