# Trigonometric relation, implication of the requirement $\theta_a = \theta_b$

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?

• The arctangent is injective on its domain. Commented Jul 4, 2019 at 18:32
• It seems to me that you need $a_1 b_1 + a_2 b_2 = 0$. If you need the coefficients to be nonnegative, then at least one of $a_1$ and $b_1$ has to be 0 and at least one of $a_2$ and $b_2$ has to be 0. Commented Jul 4, 2019 at 18:38
• @SohomPaul How do you obtain the relation $a_1 b_1 + a_2 b_2 = 0$? The coefficients $a,b$ may also be negative; there is no limitation on their sign. Commented Jul 4, 2019 at 20:03

$$A = A' \sin (n \alpha + \theta) = A' \sin(n \alpha)\cos(\theta) + A'\cos(n \alpha)\sin(\theta) = a_1 \sin(n \alpha) + a_2 \cos(n \alpha)$$
$$B = B' \cos (n \alpha + \theta) = B' \sin(n \alpha)\cos(\theta) + B'\cos(n \alpha)\sin(\theta) = b_1 \sin(n \alpha) + b_2 \cos(n \alpha)$$
$$A'\cos(\theta)= a_1$$ $$A'\sin(\theta)= a_2$$ $$B'\cos(\theta)= b_1$$ $$B'\sin(\theta)= b_2$$
$$\cos(\theta) = \dfrac{a_1}{A'} = \dfrac{b_1}{B'}$$ $$\sin(\theta) = \dfrac{a_2}{A'} = \dfrac{b_2}{B'}$$ $$A' = \sqrt{a_1^2 + a_2^2}$$ $$B' = \sqrt{b_1^2 + b_2^2}$$