Happy 2025! This math equation is finally true. When is the next time it will be true?

Anyone who has studied complex numbers, knows that De Moivre's theorem tells us that $$(\color{red}{\cos}\theta+i\color{blue}{\sin}\theta)^{2025}\equiv\color{red}{\cos}(2025\theta)+i\color{blue}{\sin}(2025\theta)$$

But

$$(\color{blue}{\sin}\theta+i\color{red}{\cos}\theta)^{2025}\equiv\color{blue}{\sin}(2025\theta)+i\color{red}{\cos}(2025\theta)$$

is also true.

The second kind of equation is not true for every year. For example, it doesn't work with 2024, or 2023.

When is the next time this kind of equation will be true?

Rewriting the base of the left hand side of the second statement, $$\color{blue}\sin(\theta) + i\color{red}\cos(\theta)= \color{red}\cos(\pi/2 - \theta) + i\color{blue}\sin(\pi/2 - \theta) = e^{i(\pi/2 - \theta)} $$

Raising this to some power $n$ is simple now, $$(\color{blue}\sin(\theta) + i\color{red}\cos(\theta))^n = e^{in(\pi/2 - \theta)} = \color{red}\cos(n\pi/2 - n\theta) + i\color{blue}\sin(n\pi/2 - n\theta)$$

For the $\color{red}\cos(n\pi/2 - n\theta)$ to become $\color{blue}\sin(n\theta)$ now and vice versa, we need $n\pi/2 \text{ mod } 2\pi = \pi/2$, or $$\frac{(n-1)\pi}{2} = 2k\pi \implies n = 4k + 1$$ for any integer $k$.

So, the next number after $2025$ to satisfy this property will be $4$ more than it, meaning the next time the condition is true is in $\mathbf{2029}$.

Another solution:

$\begin{align}\color{green}{(\sin\theta+i\cos\theta)^n}&\equiv[i(\cos\theta-i\sin\theta)]^n\\&\equiv i^n(\cos(-\theta)+i\sin(-\theta))^n\\&\equiv i^n(\cos(-n\theta)+i\sin(-n\theta))\\&\equiv i^n(\cos(n\theta)-i\sin(n\theta))\\&\equiv \color{red}{i^{n-1}}\color{green}{\left(\sin(n\theta)+i\cos(n\theta)\right)}\\\end{align}$ So if and only if $n=4k+1$, then $\color{green}{(\sin\theta+i\cos\theta)^n}\equiv \color{green}{\left(\sin(n\theta)+i\cos(n\theta)\right)}$

So the answer is:

$2029$