2009/019) Pythagorean triplets.- trigonometric way

There is a trigonometric way of getting Pythagorean triplets.

we know x^2+y^2 = z^2

is homogeneous expression and dividing it by z^2 we get

(x/z)^2 + (y/z)^2 = 1 or a^2+b^2 =1

This is an identity when we put a = sin t and b = cos t

In case we chose a and b both rational we are through. But how to we guaranty that, that is how to choose both sin t and cos t to be rational. There is a way out.

That is by using double angle formula( writing sin t and cos t in terms of tan t/2)

We know sin t = 2 sin t/2 cos t/2
= 2 tan t/2 cos ^2 t/2
= 2 (tan t/2)/(1+ tan ^2 (t/2))

And cos t = (2 cos^2(t/2) – 1) = 2/(1+ tan ^2 t/2) – 1 = (1- tan ^2 t/2)/(1+ tan ^2 t/2)

So instead of chosing 2 expressions that is sin t and cos t we can chose tan t/2 as rational and so sin t and cos t both come out to be rational

Let tan t/2 = m and we get

(2m/(1+m^2)), (1-m^2)/(1+m^2) and 1 satisfy the condition and so

2m, (1-m^2) and (1+m^2) satisfy the condition.(a^2+b^2 = c^2)

But m is not necessarily an integer and by choosing m = p/q and multiplying by q^2 we get

2pq, (q^2-p^2) and (q^2+p^2) satisfy the equation and if p and q are relatively prime then we get triplet with no common factor and hence it is a primitive triple..

Multiply this by any other integer and we shall get another triple(not primitive)