Category Archives: Year 11 Specialist Mathematics

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

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

Converting recurring (non-terminating) decimals to fractions

The easiest approach is to jump right in with some examples.

Example 1

Convert 0.\overline{5} to a fraction.

Let x=0.\overline{5}

(1)   \begin{equation*}x=0.\overline{5}\end{equation*}

(2)   \begin{equation*}10x=5.\overline{5}\end{equation*}

Subtract equation 1 from equation 2

    \begin{equation*}9x=5\end{equation}

Hence x=\frac{5}{9} so 0.\overline{5}=\frac{5}{9}

Example 2

Convert 0.\overline{12} to a fraction.

Let x=0.\overline{12}

(3)   \begin{equation*}x=0.\overline{12}\end{equation*}

(4)   \begin{equation*}100x=12.\overline{12}\end{equation*}

Subtract equation 3 from equation 4.

    \begin{equation*}99x=12\end{equation}

    \begin{equation*}x=\frac{12}{99}=\frac{4}{33}\end{equation}

Example 3

Convert 0.1\overline{23} to a fraction

Let x=0.1\overline{23}

(5)   \begin{equation*}x=0.1\overline{23}\end{equation*}

(6)   \begin{equation*}10x=1.\overline{23}\end{equation*}

(7)   \begin{equation*}1000x=123.\overline{23}\end{equation*}

Subtract equation 6 from equation 7

    \begin{equation*}990x=122\end{equation}

    \begin{equation*}x=\frac{122}{990}=\frac{61}{495}\end{equation}

Our aim is to manipulate the recurring decimal to create two numbers each which have only the repeated digits after the decimal point.

One more example.

Example 4

Convert 3.4\overline{56} to a fraction

Let x=3.4\overline{56}

If I multiply by 10, then I will have 34.\overline{56} – only repeated digits after the decimal point.

If I multiply by 1000, then I will have 3456.\overline{56}– only repeated digits after the decimal point.

So I get,

    \begin{equation*}990x=3422\end{equation}

    \begin{equation*}x=\frac{3422}{990}=3\frac{226}{495}\end{equation}

You can also use your Casio classpad to do the conversion. Although I think it is easier just to do it yourself.

Let’s think about example 4,

3.4\overline{56}=3.4+\frac{56}{1000}+\frac{56}{100000}+\frac{56}{10000000}+...

Which is

3.4+\frac{56}{1000\times 100^0}+\frac{56}{1000\times100^1}+\frac{56}{1000\times 100^2}...

3.4+\Sigma_{x=0}^\infty(\frac{56}{1000\times 100^x})

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Filed under Arithmetic, Decimals, Fractions, Year 11 Specialist Mathematics

A Counting Question

How many three digit numbers can you form from the digits 1, 2, 3, 4 and 5 if

(a) the digits must occur in increasing order?

(b) adjacent digits differ by 2?

Cambridge Year 11 Specialist Mathematics Skill Sheet 1A

(a) There are 5\times 4\times 3=60 permutations of three digits from the five digits, but how many of those are in the right order?

Each set of 3 digits has 6 arrangements (3\times 2\times 1=6).

For example, if the set is {1, 2, 3}, then the possible arrangements are:

123, 132, 213, 231, 312, and 321.

Only one of those arrangements is in numerical order.

Hence what we want is \begin{pmatrix}5\\3\end{pmatrix}=10

(b) I think this one is more about creating a list. I shall start with 1

135, 531, 131, 242, 353, 424, 535, 315

There are 8 possibilities.

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Filed under Counting Techniques, Year 11 Specialist Mathematics

Circle Geometry Problem

In the diagram, points A, B, C, D and Q lie on a circle centre O, radius 6 cm and diameter BQ, \angle{ABQ}=50^\circ, AB is parallel to DO and point P lies on diameter BQ such that OP=DP=4cm.

(a) Find \angle{BCD}

(b) Determine the length of PC.

\angle{BOD}=180^\circ-50^\circ=130^\circ (Co-interior angles in parallel lines are supplementary.)

\angle{BCD}=\frac{1}{2}\times 130=65^\circ (Angles subtended by the same arc. The angle at the centre is twice the angle at the circumference.)

\angle{BCD}=65^\circ.

Let PC=x

From the intersecting chord theorem

    \begin{equation*}4\times x=2\times 10\end{equation}

    \begin{equation*}4x=20\end{equation}

    \begin{equation*}x=5\end{equation}

PC=5cm


A chord AB of a circle O is extended to C. The straight line bisecting \angle{OAB} meets the circle at E. Let \angle{BAE}=x. Prove that EB bisects \angle{OBC}.

\angle{BAO}=2x (AE bisects \angle {BAO})

\Delta AOB is isosceles (AO=B0 radii of the circle)

\angle{ABO}=2x (Equal angles in isosceles triangle)

Therefore \angle {AOB}=180^\circ-4x (angle sum of a triangle)

\angle {BEA}=90^\circ-2x (angle at the circumference is half angle at the centre)

\angle{ABE}=180^\circ-x-(90^\circ-2x)=90^\circ+x (angle sum of a triangle)

\angle{CBE}=180^\circ-(90^\circ+x)=90^\circ-x (angles on a straight line)

\angle{OBE}=90^\circ+x-2x=90^\circ-x

\angle{OBE}=\angle{CBE}

Hence, BE bisects \angle{OBC}

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Filed under Circle Theorems, Geometry, Uncategorized, Year 11 Specialist Mathematics

Vector and Scalar Projection

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

Using a Vector Method to Find an Angle Bisector

Points A and B are defined by the position vectors \mathbf{a}=3\mathbf{i}+4\mathbf{j} and \mathbf{b}=12\mathbf{i}+5\mathbf{j}.

Find a vector that bisects \angle{AOB}.

If we think about how we add vectors using the parallelogram rule

Finding the resultant vector using the parallelogram rule

we can take advantage of the geometric properties of parallelograms (or of a rhombus).

If \mathbf{a} and \mathbf{b} are unit vectors, then the parallelogram is a rhombus, and the diagonal (i.e the resultant) bisects the angle.

We need to find the sum of the unit vectors.

|\mathbf{a}|=\sqrt{3^2+4^2}=5

\therefore \hat{\mathbf{a}}=\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}

|\mathbf{b}|=\sqrt{12^2+5^2}=13

\therefore \hat{\mathbf{b}}=\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}

The vector that bisects \angle{AOB} is

\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j}+\frac{12}{13}\mathbf{i}+\frac{5}{13}\mathbf{j}=\frac{99}{65}\mathbf{i}+\frac{64}{65}\mathbf{j}

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Filed under Geometry, Vectors, Year 11 Specialist Mathematics

Intersecting Secant Theorem

CD is a tangent to the circle.

Prove c^2=a(a+b)

I am going to add two chords to the circle

Chord AD and BD are added

\angle{BDC}=\angle{CAD} (angles in alternate segments are congruent)

\angle{BCD}=\angle{DCA} (shared angle)

\therefore \Delta BDC\cong \Delta{DAC} (AA)

Hence

\frac{DC}{AC}=\frac{BC}{DC} (Corresponding sides in similar triangles)

\frac{c}{a+b}=\frac{a}{c}

\therefore c^2=a(a+b)

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Filed under Circle Theorems, Geometry, Year 11 Specialist Mathematics

Circle Geometry Question 2

One of my Year 11 Specialist students had this question

Triangle ABC touches the given circle at Points P, Q, R and S only. The secant BW touches the circle at V and W.

Diagram not drawn to scale

(a) Determine the lengths of the line segments marked x, y and z, leaving your answers as exact values.

(b) If the length of the line segment QW is 4 units, determine the exact radius of the circle.

(a) We are going to use the Intersecting Secant Theorem – the tangent version

c^2=a\times(a+b)

Hence, we have

    \begin{equation*}30^2=25(25+x+6)\end{equation}

    \begin{equation*}900=25(31+x)\end{equation}

    \begin{equation*}x=5\end{equation}

Then we can use the intersecting chord theorem to find y.

    \begin{equation*}10\times y=6 \times x\end{equation}

    \begin{equation*}10y=30\end{equation}

    \begin{equation*}y=3\end{equation}

Back to the Intersecting Secant Theorem to find z

    \begin{equation*}z^2=4\times 17\end{equation}

    \begin{equation*}z=2\sqrt{17}\end{equation}

(b)


QW is part of a 3-4-5 triangle, therefore \angle{Q}=90^\circ

This is definitely the case of the diagram not being drawn to scale. If \angle{Q}=90^\circ, then the purple line must be the diameter.

We can use pythagoras to find the length of the diameter

    \begin{equation*}(2r)^2=13^2+4^2\end{equation}

    \begin{equation*}4r^2=185\end{equation}

    \begin{equation*}r=\frac{\sqrt{285}}{2}\end{equation}

The radius of the circle is \frac{\sqrt{285}}{2}

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Filed under Circle Theorems, Geometry, Pythagoras, Year 11 Specialist Mathematics

Circle Geometry Question

In the above diagram O is the centre of the larger circle. A, B,D and E are points on the circumference of the larger circle. A, C, E and 0 are points on the circumference of the smaller circle. Show that \angle{CAB}=\angle{ABC}. AB, AC and BC are straight lines.

AO=OB (radii of the larger circle)

At a line from O to E (it is also a radius of the larger circle)

Let \angle{CAB}=\alpha.

ACEO is a cyclic quadrilateral.

Hence, \angle{CED}=180-\alpha (AECO is a cyclic quadrilateral)

As CB is a straight line \angle{OEB}=180-(180-\alpha)=\alpha.

\Delta OEB is an isosceles triangle.

Therefore, \angle{ABC}=\alpha

Therefore \angle{ABC}=\angle{CAB}

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