I have a question that defines a few sets, then asks to prove that T is a subset of C, using the Pythagorean identity. The problem is as follows:
$P = \left \{ {\frac{a}{c} \;|\; a,c\in \mathbb{N} \; and\; a^2+b^2=c^2 \;for\;some\; b\in \mathbb{N}}\right \}$
$S = \left \{ x\in \mathbb{R} \; |\; sin(x)\in \mathbb{Q}\right \}$
$T = \left \{ x\in \mathbb{R} \; |\; sin(x)\in P\right \}$
$C = \left \{ x\in \mathbb{R} \; |\; cos(x)\in \mathbb{Q}\right \}$
Prove that $T\subseteq C$
So far I have determined that:
$cos(x) = \frac{b}{c}$
$sin(x) = \frac{a}{c}$
$\left ( \frac{a}{c} \right )^2+\left ( \frac{b}{c} \right )^2 = 1$ so
$a^{2}+b^2=c^2$
Which is pretty obvious but I'm still stuck on how to prove this subset. Any help would be greatly appreciated!
You are basically on the right track with what you've done so far to prove $T\subseteq C$. Choose any $x \in T$. This means
$$\sin(x) = \frac{a}{c} \tag{1}\label{eq1A}$$
for some $a,c \in N$ such that for some $b \in N$ you have
$$a^2 + b^2 = c^2 \implies c^2 - a^2 = b^2 \tag{2}\label{eq2A}$$
You also have from $\sin^2(x) + \cos^2(x) = 1$, plus from \eqref{eq1A} and \eqref{eq2A}, that
$$\begin{equation}\begin{aligned} \cos(x) & = \pm\sqrt{1 - \sin^2(x)} \\ & = \pm\sqrt{1 - \frac{a^2}{c^2}} \\ & = \pm\sqrt{\frac{c^2 - a^2}{c^2}} \\ & = \pm\sqrt{\frac{b^2}{c^2}} \\ & = \pm\left(\frac{b}{c}\right) \end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Since $b$ and $c$ are integers, this means $\cos(x) \in \mathbb{Q}$ and, thus, $x \in C$. Since $x$ was any element of $T$, this shows all elements of $T$ are in $C$ and, thus, $T$ is a subset of $C$, i.e.,
$$T\subseteq C \tag{4}\label{eq4A}$$
as was asked to be proven.