When taking components of a vector, why do we use trigonometric projections $(\cos θ, \sin θ)$?

Question

This question is very fundamental, but I am not being able to make sense of it. Can somebody please explain the mathematics of it.

Answer 1

Because the projection of a vector produces a right triangle when decomposed along two orthogonal directions

The vector $\vec{r}$ is decomposed as $$\vec{r} = \vec{x} + \vec{y}$$

and the sides of the triangle drawn above are found using the trigonometric relationships

$$ \begin{aligned} \cos \theta & = \frac{x}{r} & \sin \theta & = \frac{y}{r} \end{aligned} $$

Answer 2

The reason is that we define specific basis vectors. We write a vector as the sum of 2 perpendicular basis vectors, inline with a specific chosen coordinate system.

I assume you understand that if we draw a triangle, with an angle theta, the opposite and adjacent sides are given as the magnitude multiplied by either a $\sin{\theta}$ or a $\cos{\theta}$

This is just basic trigonometry which comes from the definition of sin/cos.

I feel like the missing connection in answering your question, is:

Which angle of theta do I use?, if I take a vector, I could draw INFINITELY many triangles with that vector, all with varying angles of theta.

The triangle we "define" to represent the components of our vector, is the triangle whose sides point in the direction of our CHOSEN basis vectors. These lengths are then the component of length purely in the direction of the chosen basis vectors, whose sum is therefore the actual vector.