rational sine and cosine values such that the numerators are odd.

Question

Consider the angle $\theta$ with $$0<\theta<\frac{\pi}{2}$$ Suppose that $$cos\theta=\frac{l}{n} \space and \space \space sin\theta=\frac{m}{n}$$ are rational numbers, with postive integers $l,m$ and $n$. Show that $l$ and $m$ cannot both odd integers.
MY ATTEMPT: Assume that $l$ and $m$ are both odd integers. Then $cos^2\theta=\frac{l^2}{n^2}$ and $sin^2\theta=\frac{m^2}{n^2}$ then adding these equations and using the Pythagorean identity we obtain $$l^2+m^2=n^2$$ Notice we can conclude n is even, and we must find a contradiction. I am unsure where to go from here, any help is appreciated!

Answer 1

You are on the right track. Following on from your proof:

let $m = 2k_{1}+1, l=2k_{2} +1, n = 2k_{3}$

$m^2 + l^2 = n^2\\ $ $\Rightarrow (2k_1+1)^2 + (2k_2+1)^2 = (2k_3)^2\\ $

$\Rightarrow 4k_1^2 + 4k_1 + 1 + 4k_2^2 + 4k_2 + 1 = 4k_3^2\\ $

$\Rightarrow 4(k_1^2 + k_1 + k_2^2 + k_2) + 2 = 4(k_3^2) $

Which is clearly a contradiction since we have ($4\cdot$ integer )$+2$ = $4\cdot$ integer

Thus $m$ and $l$ cannot both be odd.

Answer 2

Hint: Assume they are, write them as $l=(2k+1)$ and $m=(2j+1)$, and expand to see what n has to be equal to:

$l^2+m^2 = (2k+1)^2+(2j+1)^2 = 4k(k+1)+4j(j+1)+2$