Multiplying both quantities, we get $\sin x \cos x + \sin x + \cos x = r\in\mathbb{Q}$ for some $r$. Now, let $c=\sin 2x = 2\sin x \cos x$. We then get \[ \frac{c}{2} \pm \sqrt{1+2c} = r \Rightarrow \frac{c^2}{4}-cr + r^2 = 1+2c. \]So, $c$ is a root of a quadratic trinomial with rational coefficients. Multiplying both sides by $4N^2$, where $N\in\mathbb{N}$ with $Nr\in\mathbb{Z}$, we immediately obtain the desired result.

grupyorum wrote: Multiplying both quantities, we get $\sin x \cos x + \sin x + \cos x = r\in\mathbb{Q}$ for some $r$. Now, let $c=\sin 2x = 2\sin x \cos x$. We then get \[ \frac{c}{2} \pm \sqrt{1+2c} = r \Rightarrow \frac{c^2}{4}-cr + r^2 = 1+2c. \]So, $c$ is a root of a quadratic trinomial with rational coefficients. Multiplying both sides by $4N^2$, where $N\in\mathbb{N}$ with $Nr\in\mathbb{Z}$, we immediately obtain the desired result. Sir, how did you get the $\pm \sqrt{1+2c}$ term?

RenheMiResembleRice wrote: grupyorum wrote: Multiplying both quantities, we get $\sin x \cos x + \sin x + \cos x = r\in\mathbb{Q}$ for some $r$. Now, let $c=\sin 2x = 2\sin x \cos x$. We then get \[ \frac{c}{2} \pm \sqrt{1+2c} = r \Rightarrow \frac{c^2}{4}-cr + r^2 = 1+2c. \]So, $c$ is a root of a quadratic trinomial with rational coefficients. Multiplying both sides by $4N^2$, where $N\in\mathbb{N}$ with $Nr\in\mathbb{Z}$, we immediately obtain the desired result. Sir, how did you get the $\pm \sqrt{1+2c}$ term? $\sin x+\cos x=\pm \sqrt{{{\left( \sin x+\cos x \right)}^{2}}}=\pm \sqrt{{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x}=\pm \sqrt{1+2\sin x\cos x}=\pm \sqrt{1+2c}$

phantomofaops wrote: RenheMiResembleRice wrote: grupyorum wrote: Multiplying both quantities, we get $\sin x \cos x + \sin x + \cos x = r\in\mathbb{Q}$ for some $r$. Now, let $c=\sin 2x = 2\sin x \cos x$. We then get \[ \frac{c}{2} \pm \sqrt{1+2c} = r \Rightarrow \frac{c^2}{4}-cr + r^2 = 1+2c. \]So, $c$ is a root of a quadratic trinomial with rational coefficients. Multiplying both sides by $4N^2$, where $N\in\mathbb{N}$ with $Nr\in\mathbb{Z}$, we immediately obtain the desired result. Sir, how did you get the $\pm \sqrt{1+2c}$ term? $\sin x+\cos x=\pm \sqrt{{{\left( \sin x+\cos x \right)}^{2}}}=\pm \sqrt{{{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x}=\pm \sqrt{1+2\sin x\cos x}=\pm \sqrt{1+2c}$ After squaring, where does the the -cr term come from?