Displacement, velocity, acceleration and force all have a size and a direction: they are vectors. In two dimensions the easiest way to handle a vector is to split it into an $x$-part and a $y$-part, using the Sine, Cosine and Tangent you know from geometry.
Key idea
A vector of magnitude $A$ at angle $\theta$ from the $+x$ axis has components $A_x = A\cos\theta$ and $A_y = A\sin\theta$. In general: the component along the side adjacent to the angle uses cosine; the one opposite uses sine.
Why it works
Draw the vector as the hypotenuse of a right triangle whose legs lie along the axes. The legs are the components. Cosine is adjacent over hypotenuse, so the adjacent leg is $A\cos\theta$; sine is opposite over hypotenuse, so the opposite leg is $A\sin\theta$. Which leg is "adjacent" depends on which axis the angle is measured from, not on whether it is $x$ or $y$.
Worked examples
Worked example: Angle from the horizontal
A rope pulls a sled with $80\ \text{N}$ at $25^\circ$ above the horizontal. Find the components.
- The angle is measured from $x$, so $x$ is adjacent: $F_x = 80\cos 25^\circ \approx 72.5\ \text{N}$.
- $F_y = 80\sin 25^\circ \approx 33.8\ \text{N}$.
Answer: $F_x \approx 72.5\ \text{N}$, $F_y \approx 33.8\ \text{N}$
Worked example: Angle from the vertical
A kicked ball leaves at $15\ \text{m/s}$, $20^\circ$ from the vertical, leaning toward $+x$. Find $v_x$.
- Now the angle is measured from the $y$ axis, so $y$ is adjacent and $x$ is opposite.
- $v_x = 15\sin 20^\circ \approx 5.13\ \text{m/s}$ and $v_y = 15\cos 20^\circ \approx 14.1\ \text{m/s}$.
Answer: $v_x \approx 5.13\ \text{m/s}$
Worked example: Directions with signs
A hiker's displacement is $300\ \text{m}$ at $40^\circ$ south of west. With east $+x$ and north $+y$, find the components.
- "South of west": start pointing west, turn $40^\circ$ toward south. The angle is measured from the $x$ axis (west).
- Sizes: $300\cos 40^\circ \approx 230\ \text{m}$ (along $x$) and $300\sin 40^\circ \approx 193\ \text{m}$ (along $y$).
- Signs: west is $-x$ and south is $-y$, so $d_x \approx -230\ \text{m}$ and $d_y \approx -193\ \text{m}$.
Answer: $d_x \approx -230\ \text{m}$, $d_y \approx -193\ \text{m}$
Common mistake
Using sine for the $x$-component every time (or cosine every time). It depends on where the angle is measured from. Quick checks: as $\theta \to 0$ from an axis, the component along that axis should approach the whole vector (cosine does that). And the two components must satisfy $A_x^2 + A_y^2 = A^2$.