Are $(\sin 49^{\circ})^2$ and $(\cos 49^{\circ})^2$ irrational numbers?
When you enter, $(\sin 49^{\circ})^2$ in a calculator, it shows a long number (and if it is irrational, then clearly the calculator cannot calculate that number to the last digit. i.e., it gives you an approximate for $(\sin 49^{\circ})^2$).
Now save that number in the memory of the calculator, and then calculate $(\cos 49^{\circ})^2$. Now add these numbers up. You will get $1$.
But how this happens?
I am almost sure that the numbers $\sin^2 49^{\circ}$ and $\cos^2 49^{\circ}$ are irrational, and I don't know how does the calculator gives the exact $1$ when you add these up.
Best Answer
First, the way many scientific calculators work is to calculate more digits than they show, and then they round the value before displaying it. This happens even if you calculate something like $1/7 \times 7$. The calculator may believe the result is slightly lower than $1$, but the rounded number is $1$. You can test how many digits of precision your calculator uses by multiplying by $10^n$ and then subtracting the integer part. This will often reveal a few more digits.
Second, those are irrational numbers. Proving that takes some number theory. Let $\xi = \cos 1^\circ + i \sin 1^\circ$, a $360$th root of unity. $\xi$ is conjugate to $\xi^n$ for each $n$ coprime to $360$ including $\xi^{49} = \cos 49^\circ + i \sin 49^\circ$. The minimal polynomial of $\xi$ and $\xi^{49}$ has degree $\phi(360)=96$. If $\cos^2 49^\circ$ were rational, then $(\xi^{49} + \xi^{-49})^2$ would be rational, which would mean that $\xi^{49}$ satisfies a polynomial with rational coefficients of degree $4 \lt 96$. Similarly, $\sin^2 49^\circ = (\xi^{49} - \xi^{-49})^2/4$ is not rational.