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 ).

Integration by Substitution

Current UK exam textbooks pass over proofs and mathematical discussions in a hurry to show the ‘how to’ of exam questions.

Integration by substitution is a little more than just backwards chain-rule and deserves a fuller treatment.

Try this,

Let,

y=\displaystyle \int \textnormal{f}(x) \ \textnormal{d}x

then,

\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{f}(x)

Suppose that there exists a function g, of another variable u, such that x=\textnormal{g}(u) and let, \textnormal{f}(x)=\textnormal{f}(\textnormal{g}(u))=\textnormal{F}(u). So that,

\dfrac{\textnormal{d}y}{\textnormal{d}x}=\textnormal{F}(u)

Now, by the chain rule,

\dfrac{\textnormal{d}y}{\textnormal{d}u}=\dfrac{\textnormal{d}y}{\textnormal{d}x}\times \dfrac{\textnormal{d}x}{\textnormal{d}u}=\textnormal{F}(u)\dfrac{\textnormal{d}x}{\textnormal{d}u}

Hence,

y=\displaystyle \int \dfrac{\textnormal{d}y}{\textnormal{d}u} \ \textnormal{d}u=\displaystyle \int \textnormal{F}(u)\dfrac{\textnormal{d}x}{\textnormal{d}u} \ \textnormal{d}u

i.e.

y=\displaystyle \int \textnormal{f}(x) \ \textnormal{d}x=\displaystyle \int \textnormal{F}(u)\dfrac{\textnormal{d}x}{\textnormal{d}u} \ \textnormal{d}u

FP3 2014 gives students the chance to prove the surface area of a sphere is 4 pi r squared.  Interestingly four time the area of a cross section through the centre and the derivative of the volume.  Pupils with strong maths general knowledge from GCSE would know what is coming.  TOP TIP do as much maths challenge and extension work as possible, the general knowledge thus gained can only help.  This FP3 2014 paper and many others are at my Exam Bucket: http://jped.co.uk/ExamBucket/exambucket.php.