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 ).

Area of a circle – proof by exhuastion

Once \pi has been defined as the ratio circumference to diameter, the area of a circle must be \pi r^{2}.

A proof relies on an infinite, limiting process which paves the way to some calculus-like ideas.

A circle is cut up into 6 sectors which are then rearranged into a near rectangle,

What doesn’t look too close to a rectangle at 6 sectors looks better at 26:

As the number of sectors becomes very big the shape becomes indistinguishable from a rectangle and the argument is complete.

I think that Archimedes used such arguments, or, proof by exhaustion, in some of his solid geometry work.

Parallelograms generalise Pythagoras to any triangle

There’s always space on the inter-web for another proof of Pythagoras’s Theorem.  Here’s one that uses the following equal areas property of parallelograms.

160530AreaOfParallelogram

This kind of area chopping and shape translation is a feature of Euclidean geometry and our senses support it’s veracity at the order of size of the classroom.

The squares on the sides of a right-angle triangle set up a system of parallel lines which can then be used to demonstrate the Theorem using the above equal areas property.

160529PythagProof

The thread does not stop here though.  Taking the parallel line structure which makes this work we get a generalisation of Pythagoras to non-right-angled triangles with the area of the parallelogram on the longest side being the sum of the areas of those constructed on the other two sides.

160530PythagGeneralised

Before the beginning – Euclid’s Common Notions

This preamble from Euclid’s Elements is where mathematics education still goes wrong, even after 2000 years or so.

Sound notions of what equality is are required as a bedrock of algebra.  Dynamic approaches of teaching equation solving, in which terms and numbers move and change sign, seem to work at first but lack the simplicity of mathematical logic and create all sorts of problems when randomly, but apparently sensibly, applied.