December 4, 2025 · 12:52 pm

Area of a triangle from the semi-perimeter and the radius of the incircle.

    \begin{equation*}A=sr\end{equation}

Where s is the semi-perimeter, s=\frac{a+b+c}{2} and r is the radius of the incircle.

AB, BC and AC are tangents to the circle. And the radii are perpendicular to the tangents.

Add line segments AO, CO and BO.

\Delta{ABC} is split into three triangles, \Delta{AOB}, \Delta{AOC} and \Delta{BOC}.

Hence Area \Delta{ABC}=\Delta{AOB}+\Delta{AOC}+\Delta{BOC}

\Delta{ABC}=\frac{1}{2}cr+\frac{1}{2}br+\frac{1}{2}ar

\Delta{ABC}=\frac{1}{2}r(a+b+c)

Remember s=\frac{1}{2}(a+b+c)

\Delta{ABC}=sr

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Filed under Area, Finding an area, Geometry, Interesting Mathematics, Radius and Semi-Perimeter

November 26, 2025 · 10:57 am

Trigonometric Equation

Solve cos(4\theta)+cos(2\theta)+cos(\theta)=0 for 0\le \theta \le\pi

Remember the identity

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

Hence

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

Now I have

    \begin{equation*}2cos(3\theta)cos(\theta)+cos(\theta)=0\end{equation}

    \begin{equation*}cos(\theta)(2cos(3\theta)+1)=0\end{equation}

cos(\theta)=0 or cos(3\theta)=\frac{-1}{2}

\theta=\frac{\pi}{2}

cos(3\theta)=-\frac{1}{2} for 0 \le \theta \le 3\pi

3\theta=\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}

\theta=\frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9}

Hence \theta =\frac{\pi}{2},\frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9}

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November 22, 2025 · 6:10 am

Polynomial Long Division

I usually choose to use synthetic division when factorising polynomials, but I know some teachers are unhappy when their students do this. So for completeness, here is my PDF for Polynomial Long Division.

Polynomial Long DivisionDownload

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Filed under Algebra, Cubics, Factorisation, Factorising, Factorising, Polynomials, Quadratics, Solving, Solving, Solving Equations, Year 11 Mathematical Methods

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November 16, 2025 · 5:54 am

Geometry Circle Question

In the diagram below, A, B, C and D lie on the circle with centre O. If \angle{DBC} = 41^{\circ} and \angle{ACD} = 53^{\circ}, determine with reasoning \angle{BAC} and \angle{AOB}

We know OA=OB=OD – radii of the circle.

Which means, \Delta{AOB} is isosceles and \angle{OAB}=\angle{OBA} – equal angles isosceles triangle.

\angle{AOD}=2\angle{ACD} – angle at the centre twice the angle at the circumference.

\angle{AOB}=106^{\circ}

This means \angle{AOB}=74^{\circ} – angles on a straight line are supplementary

\angle{OAD}=\angle{ODA}=37^{\circ} – equal angles isosceles triangle and the angle sum of a triangle.

\angle{DBA}=\angle{DCA}=53^{\circ} – angle at the circumference subtended by the same arc are congruent.

\angle{CAD}=\angle{CBD}=41^{\circ} – angles at the circumference subtended by the same arc are congruent.

\angle{OAB}=53^{\circ} – equal angle isosceles triangle

Hence \angle{BAC}=12^{\circ}

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Filed under Circle Theorems, Finding an angle, Geometry, Year 11 Specialist Mathematics

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November 11, 2025 · 10:19 am

Completing the Square

x^2+6x-4 \rightarrow (x+3)^2-13

Completing the square is useful to

  • sketch parabolas.
  • solve quadratics.
  • factorising quadratics
  • finding the centre and radius version of the equation of a circle.

When completing the square we take advantage of perfect squares. For example, (x+3)^2=(x+3)(x+3)=x^2+6x+9

6=2\times 3 and 9=3\times 3

Example 1

Put x^2+8x-5 into completed square form.

What perfect square has an 8x term?

(x+4)^2=x^2+8x+16

We don’t want +16, we want -5, so subtract 16+5

x^2+8x-5=(x+4)^2-21

x^2+bx+c=(x+\frac{b}{2})^2-(\frac{b}{2})^2+c

What about a non-monic quadratic? For example,

2x^2+12x+11

Factorise the 2

2(x^2+6x+\frac{11}{2})

And continue as before

2[(x+3)^2-9+\frac{11}{2}]=2[(x+3)^2-\frac{18}{2}+\frac{11}{2}]=2[(x+3)^2-\frac{7}{2}]=2(x+3)^2-7

Example 2

y=2x^2+7x-5

2(x^2+\frac{7}{2}x-\frac{5}{2})

2[(x+\frac{7}{4})^2-(\frac{7}{4})^2-\frac{5}{2}]

2[(x+\frac{7}{4})^2-\frac{49}{16}-\frac{40}{16}]

2[(x+\frac{7}{4})^2-\frac{89}{16}]

2(x+\frac{7}{4})^2-\frac{89}{8}

ax^2+bx+c=a(x+\frac{b}{2a})^2-a((\frac{b}{2a})^2+\frac{c}{a})

Casio Classpad e-activity

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Filed under Algebra, Arithmetic, Classpad Skills, Completing the Square, Fractions, Quadratic, Quadratics, Year 10 Mathematics, Year 9 Mathematics

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November 5, 2025 · 6:57 am

Equation of a Circle and Geometry Question

A circle has equation x^2+y^2+4x-6y=36
(a) Find the centre and radius of the circle.
Points S and T lie on the circle such that the origin is the midpoint of ST.
(b) Show that ST has a length of 12.

(a)We need to put the circle equation into completed square form

    \begin{equation*}(x+2)^2-4+(y-3)^2-9=36\end{equation}

    \begin{equation*}(x+2)^2+(y-3)^2=49\end{equation}

The centre is (-2, 3) and the radius is 7.

(b)Draw a diagram

We know SO and TO are radii of the circle. Hence \Delta{SOT} is isosceles and the line segment from O to the origin is perpendicular to ST.

OT=7 and the distance from O to the origin is

    \begin{equation*}\sqrt{(-2-0)^2+(3-0)^2}=\sqrt{13}\end{equation}

We can use Pythagoras to find the distance from the origin to T.

    \begin{equation*}x=\sqrt{7^2-(\sqrt{13})^2}=\sqrt{49-13}=\sqrt{36}=6\end{equation}

Hence ST=2\times6=12


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Filed under Co-ordinate Geometry, Geometry, Pythagoras, Year 11 Mathematical Methods

October 28, 2025 · 5:42 am

Integration Question (much easier with Integration by Parts)

The Year 12 Mathematics Methods course doesn’t cover Integration by Parts, so they end up with questions like the following.

Determine the following:
(a) \frac{d}{dx} \left (e^{2x}sin(3x) \right )
(b) \frac{d}{dx} \left (e^{2x}cos(3x) \right )
Hence, determine the following integral by considering both parts (a) and (b)
\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx

(a) Use the product rule

(1)   \begin{equation*}\frac{d}{dx} \left (e^{2x}sin(3x) \right)= 2e^{2x}sin(3x)+3e^{2x}cos(3x) \end{equation*}

(b)

(2)   \begin{equation*}\frac{d}{dx} \left (e^{2x}cos(3x) \right )=2e^{2x}cos(3x)-3e^{2x}sin(3x)\end{equation*}

I need to use equations 1 and 2 to find \int_0^{\frac{\pi}{2}}13e^2xcos(3x) \enspace dx.

The e^{2x}sin(3x) terms need to vanish and I need 13 of the e^{2x}cos(3x) terms.

3\times \text{equation} (1)+ 2\times \text{equation} (2)

(3)   \begin{equation*}3\frac{d}{dx} \left (e^{2x}sin(3x) \right)= 6e^{2x}sin(3x)+9e^{2x}cos(3x) \end{equation*}

(4)   \begin{equation*}2\frac{d}{dx} \left (e^{2x}cos(3x) \right )=4e^{2x}cos(3x)-6e^{2x}sin(3x)\end{equation*}

Equation 3 plus equation 4

(5)   \begin{equation*}3\frac{d}{dx} \left (e^{2x}sin(3x) \right)+2\frac{d}{dx} \left (e^{2x}cos(3x) \right )=13e^{2x}cos(3x)\end{equation*}

Integrate both sides of the equation

\int_0^{\frac{\pi}{2}} \left (3\frac{d}{dx} \left (e^{2x}sin(3x) \right)+2\frac{d}{dx} \left (e^{2x}cos(3x) \right ) \right ) dx=\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx

By the fundamental theorem of calculus, we know

(3e^{2x}sin(3x) \right)+2 \left (e^{2x}cos(3x) \right ]_0^{\frac{\pi}{2}}=\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx

3e^{\pi}sin(\frac{3\pi}{2}})+2e^{\pi}cos(\frac{3\pi}{2}})-3e^0sin(3(0))-2e^0cos(3(0))=\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx

\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx=-3e^{\pi}-2

Integration by Parts

Remember \int u\enspace dv=uv-\int v \enspace du

\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx

Let u=cos(3x), then du=-3sin(3x)

and dv=e^{2x}, then v=\frac{e^{2x}}{2}

\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx = \left (cos(3x)\left (\frac{e^{2x}}{2}\right ) \right )_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \frac{e^{2x}}{2}(-3sin(3x)) \enspace dx

\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=cos(\frac{3\pi}{2})(\frac{e^\pi}{2})-cos(0)(\frac{e^0}{2})+\frac{3}{2}\int_0^{\frac{\pi}{2}}sin(3x)e^{2x} \enspace dx

\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}+\frac{3}{2}\int_0^{\frac{\pi}{2}}sin(3x)e^{2x} \enspace dx

Let u=sin(3x), then du=3cos(3x)

and dv=e^{2x}. then v=\frac{e^{2x}}{2}

\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}+\frac{3}{2}(\frac{e^{2x}}{2}sin(3x)]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \frac{e^{2x}}{2}(3cos(3x)) \enspace dx)

\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}+\frac{3}{2}(-\frac{e^\pi}{2}-\frac{3}{2} \int_0^{\frac{\pi}{2}} e^{2x}cos(3x) \enspace dx)

\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}-\frac{3e^\pi}{4}-\frac{9}{4} \int_0^{\frac{\pi}{2}} e^{2x}cos(3x) \enspace dx

Collect like terms (the integrals are like)

\frac{13}{4}(\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx)=-\frac{1}{2}-\frac{3e^\pi}{4}

(\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx)=\frac{-2-3e^\pi}{13}

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October 20, 2025 · 11:11 am

Fundamental Theorem of Calculus

(1)   \begin{equation*}\frac{d}{dx} \left (\int_2^{f(x)} g(t) dt \right )=g(f(x))f'(x)\end{equation*}

My Year 12 Mathematics Methods students are getting ready for their exam, and questions using the above idea have created a bit of consternation. I am going to work through an example, and show why the ‘formula’ works.

Example

Find \frac{d}{dx} \left ( \int_2^{x^2} 4t^2+3t \enspace dt \right ).

    \begin{equation*}=\frac{d}{dx} \left (\frac{4t^3}{3}+\frac{3t^2}{2} \right ]_2^{x^2}\end{equation}

    \begin{equation*}=\frac{d}{dx} \left ( \frac{4(x^2)^3}{3}+\frac{3(x^2)^2}{2} -(\frac{4(2^3)}{3}+\frac{3(2^2)}{2} \right )\end{equation}

    \begin{equation*}=\frac{d}{dx} \left (\frac{4x^6}{3}+\frac{3x^4}{2}-\frac{50}{3} \right) \end{equation}

    \begin{equation*}=\frac{24x^5}{3}+\frac{12x^3}{2}\end{equation}

(2)   \begin{equation*}=8x^5+6x^3\end{equation*}

If we used ‘formula’ 1

    \begin{equation*}\frac{d}{dx} \left ( \int_2^{x^2} 4t^2+3t \enspace dt \right )=(4(x^2)^2+3(x^2))(2x)\end{equation}

\

(3)   \begin{equation*}\frac{d}{dx} \left ( \int_2^{x^2} 4t^2+3t \enspace dt \right )=8x^5+6x^3 \end{equation*}

We can see equation 2 and 3 are the same.

More formally

    \begin{equation*}\frac{d}{dx} \left ( \int_c^{f(x)} g(t) dt \right )=\frac{d}{dx} \left (G(f(x))-G(c) \right ) \end{equation}

Remember \frac{d}{dx} \left (f(g(x)) \right )=f'(g(x))\times g'(x)

    \begin{equation*}\frac{d}{dx} \left (G(f(x))-G(c) \right ) =G'(f(x))f'(x)=g(f(x))\times f'(x)\end{equation}

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October 15, 2025 · 6:19 am

Hard Equation Solving Question

Find the value(s) of k such that the equation below has two numerically equal but opposite sign solutions (e.g. 5 and -5).

    \begin{equation*}\frac{x^2-2x}{4x-1}=\frac{k-1}{k+1}\end{equation}

    \begin{equation*}(x^2-2x)(k+1)=(k-1)(4x-1)\end{equation}

    \begin{equation*}(k+1)x^2-2kx-2x=4kx-k-4x+1\end{equation}

    \begin{equation*}(k+1)x^2-2kx-4kx-2x+4x-1=0\end{equation}

    \begin{equation*}(k+1)^2x^2-(6k-2)x-1=0\end{equation}

For there to be two numerically equal but opposite sign solutions, the b term of the quadratic equation must be 0.

    \begin{equation*}6k-2=0\end{equation}

Hence k=\frac{1}{3}.

When k=\frac{1}{3} the equation becomes

    \begin{equation*}\frac{x^2-2x}{4x-1}=\frac{\frac{-2}{3}}{\frac{4}{3}}\end{equation}

    \begin{equation*}\frac{x^2-2x}{4x-1}=\frac{-1}{2}\end{equation}

    \begin{equation*}2x^2-4x=-4x+1\end{equation}

    \begin{equation*}2x^2-1=0\end{equation}

    \begin{equation*}x^2=\frac{1}{2}\end{equation}

    \begin{equation*}x=\pm \frac{1}{\sqrt{2}}\end{equation}

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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

Useful identities
sin(2x)=2sin(x)cos(x)
cos(2x)=cos^2(x)-sin^2(x)=1-2sin^2(x)=2cos^2(x)-1
sin(3x)=3sin(x)-4sin^3(x)
cos(3x)=4cos^3(x)-3cos(x)
sin(A-B)=sin(A)cos(B)-sin(B)cos(A)
cos^2(x)=1-sin^2(x)

    \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}

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

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