This question is related to a previous one. Consider the following two quantities:
$$A = a_1 \sin (n \alpha) + a_2 \cos (n \alpha)\\ B = b_1 \sin (n \alpha) + b_2 \cos (n \alpha)$$
with
$$n \in \mathbb{N}\\ \alpha \in [0; 2\pi)\\ a_{1,2}, b_{1,2} \in \mathbb{R}$$
It is possible to rewrite them as
$$A = A' \sin (n \alpha + \theta_a)\\ B = B' \cos (n \alpha + \theta_b)$$
If I require $\theta_a = \theta_b$, does this imply any condition upon $a_{1,2}, b_{1,2}$? Note that there is no requirement on $A'$ and $B'$: they can differ.
I have no clue about how to proceed. Taking the case $a_1, a_2 > 0$, $b_1, b_2 > 0$,
$$\theta_a = -\arctan \left( \frac{a_1}{a_2} \right)$$ $$\theta_b = \arctan \left( \frac{b_2}{b_1} \right)$$
This seems to be true only when $a_1 = b_2 = 0$, or $a_1 = b_2$ and $a_2 = -b_1$. But this would violate the initial hypothesis $a_1, a_2 > 0$, $b_1, b_2 > 0$, so maybe only $a_1 = b_2 = 0$ is acceptable.
How to consider all the remaining cases?
$A = A' \sin (n \alpha + \theta) = A' \sin(n \alpha)\cos(\theta) + A'\cos(n \alpha)\sin(\theta) = a_1 \sin(n \alpha) + a_2 \cos(n \alpha)$
$B = B' \cos (n \alpha + \theta) = B' \sin(n \alpha)\cos(\theta) + B'\cos(n \alpha)\sin(\theta) = b_1 \sin(n \alpha) + b_2 \cos(n \alpha)$
You must have
$$A'\cos(\theta)= a_1$$ $$A'\sin(\theta)= a_2$$ $$B'\cos(\theta)= b_1$$ $$B'\sin(\theta)= b_2$$
Which is equivalent to
$$\cos(\theta) = \dfrac{a_1}{A'} = \dfrac{b_1}{B'}$$ $$\sin(\theta) = \dfrac{a_2}{A'} = \dfrac{b_2}{B'}$$ $$A' = \sqrt{a_1^2 + a_2^2}$$ $$B' = \sqrt{b_1^2 + b_2^2}$$