t-Formulae and parameterisation of the circle

t-Formulae are used in integration to tackle rational expressions of tigonometric functions.  After a spell in the cold, when they were not included in some A level specifications, they are now back in sixth form lessons.

It all starts with the subsitution,

t=\tan \frac{\theta}{2}

from which the following functions can be derived,

\tan \theta=\dfrac{2t}{1-t^{2}}\cos \theta =\dfrac{1-t^{2}}{1+t^{2}}\sin \theta=\dfrac{2t}{1+t^{2}}.

These derivations can be made using compound trigonmetry fomulae.  Alternatively, there is an engaging co-ordinate geometry derivation which has the merits of doubling up as an algebraic parametrisation of the circle.

Start with the following figure:

t-formulae2

Euclid tells us that the angle subtended by the chord PP' at the centre is twice the angle subtended at the circumference.  The X-axis providing a line of symmetry, gives the relationship between the angles \frac{\theta}{2} and \theta at A and O respectively.

Defining t=\tan \theta and creating a line, y=t(x+1), through A with gradient t gives intersections with the Y-axis and the circle at R and P respectively.

The intersection point P can then be found by solving the simulataneous equations:

y=t(x+1), and, x^2+y^2=1.

Substituting for y leads to the quadratic,

(t^2+1)x^2+2t^2x+t^2-1=0,

which admits an easy factorisation once one acknowledges that it must have one root of x=-1.

(x+1)(x-\frac{1-t^2}{1+t^2})=0,

giving the other root, and x value for P as \frac{1-t^2}{1+t^2}.

Solving for y gives the co-ordinates of P.  When viewed as two altenative parameterisations of the unit circle, the derivation of the t formulae is complete.

\left(\dfrac{1-t^{2}}{1+t^{2}},\dfrac{2t}{1+t^{2}}\right)=(\cos \theta,\sin \theta ).

Counting Pennies, Generating Functions and STEP

STEP I 1997, Question 1
Show that you can make up 10 pence in eleven ways using 10p, 5p, 2p and 1p coins.

In how many ways can you make up 20 pence using 20p,10p 5p, 2p and 1p coins?

There is an interesting way of approaching this question using generating functions.

A generating function is of the form:

a_{0}+a_{1}t+a_{2}t^{2}+a_{3}t^{3}+\dots

where the coefficients of t count various problems and issues of convergence are unimportant.

Show that you can make up 10 pence in eleven ways using 10p, 5p, 2p and 1p coins.

This will be the co-efficient of t^{10} in,

(1+t+t^{2}+t^{3}+\dots+t^{10})(1+t^{2}+t^{4}+\dots+t^{10})(1+t^{5}+t^{10})(1+t^{10})

by way of explanation:

\overset{\textnormal{the 1p s}}{\overbrace{(1+t+t^{2}+t^{3}+\dots+t^{10})}}\overset{\textnormal{the 2p s}}{\overbrace{(1+t^{2}+t^{4}+\dots+t^{10})}}\overset{\textnormal{the 5p s}}{\overbrace{(1+t^{5}+t^{10})}}\overset{\textnormal{the 10p s}}{\overbrace{(1+t^{10})}}

The candidate still needs to expand the function to get the answer, but having changed the problem to one of algebra, they might enjoy the benefits of practice and familiarity in the manual computation.

For those of at liberty to explore the question at leisure, a Computer Algebra System (CAS) is useful. In fact, the website Wolfram Alpha will expand the brackets for us, computing the answers to all similar counting problems in the process. The Wolfram Alpha query box understands the latex code if this is just pasted in, but only if it is presented in an explicit form, that is without the dots.  In order to simplify the question for the CAS notice that each bracket is a geometric progression, which can be summed using standard A level theory reducing the function to:

\dfrac{(1-t^{11})}{(1-t)}\dfrac{(1-t^{12})}{(1-t^{2})}\dfrac{(1-t^{15})}{(1-t^{5})}\dfrac{(1+t^{10})}{1}

Use of Wolfram Alpha then gives the following:

190218GenratingPennies1

which contains our answer to the first part: 11.

Equipped with some methods and machinery we can now proceed to the second part of the question.  In fact, we can adapt the method and functions to count any similar loose change problem.

In how many ways can you make up 20 pence using 20p,10p 5p, 2p and 1p coins?

This will be the coefficient of t^{20} in,

(1+t+t^{2}+t^{3}+\dots+t^{20})(1+t^{2}+t^{4}+\dots+t^{20})(1+t^{5}+t^{10}+\dots+t^{20})(1+t^{10}+t^{20})(1+t^{20})

\overset{\textnormal{the 1 p s}}{\overbrace{(1+t+t^{2}+t^{3}+\dots+t^{20})}}\overset{\textnormal{the 2 p s}}{\overbrace{(1+t^{2}+t^{4}+\dots+t^{20})}}\overset{\textnormal{the 5 p s}}{\overbrace{(1+t^{5}+t^{10}+\dots+t^{20})}}\overset{\textnormal{the 10p s}}{\overbrace{(1+t^{10}+t^{20})}}\overset{\textnormal{the 20p s}}{\overbrace{(1+t^{20})}}

which is equal to

\dfrac{(1-t^{21})}{(1-t)}\dfrac{(1-t^{22})}{(1-t^{2})}\dfrac{(1-t^{25})}{(1-t^{5})}\dfrac{(1+t^{10}+t^{20})(1+t^{20})}{1}

and which Wolfram Alpha gives and expansion of:

190218GenratingPennies2

and our answer: 41.