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:

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.

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

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