# When do I use the cosine and sine functions when graphing the components of a vector?

I have a vector $$F_e$$ and I would like to graph it to its corresponding $$X$$ and $$Y$$ components. I know that the $$i_y$$ component is negative, and there is no $$i_x$$ component.

$$\vec{F_e} = F_x\vec{i_x} + F_y\vec{i_y}$$

$$\vec{F_e} = F_y\vec{(-i_y)}$$

And to get there we have:

$$F_{ex}=|\vec{F_{ex}}|\cos(270^\circ)$$

$$F_{ey}=|\vec{F_{ey}}|\sin(270^\circ)$$

My question is when do we use the $$\sin(\cdot)$$ and when the $$\cos(\cdot)$$ to find its magnitude. Intuitively $$\cos (270^\circ)$$ equals $$0$$ and $$\sin (270^\circ)$$ equals $$-1$$. But why didn't we set $$F_{ex}$$ with $$\sin (270^\circ)$$. I know that the angle between the $$X$$ axis and the vector is $$270^\circ$$.

If a point is at a unit distance from the origin, and it makes an angle $$p$$ with the x-axis, then $$\cos p$$ is defined as the x co-ordinate of that point. $$\sin p$$ is defined as the y co-ordinate of that point.

If a point is a distance $$r$$ from the origin and makes an angle $$p$$ wih the x-axis, then its x and y co-ordinates are $$r \cos p$$ and $$r\sin p$$ respectively.

Since you have to find the x-component of the force here, you have to use $$|\vec{F}| \cos p$$

• This is the thing i was bothered with, it seems simple but why is cos defined with x and sin with y ? Apr 21 at 5:05
• @MarkoMajstorovic Historically, the x co-ordinate has been named cos, and the y has been named sin. Naming is arbitrary. Apr 21 at 5:39
• @MarkoMajstorovic You can think of $\cos {p}$ to mean the sentence: "The x co-ordinate of the point which is at a unit distance from the origin, and makes an angle $p$ with the x-axis". The name $\cos{p}$ has been assigned to this sentence. Apr 21 at 5:43
• @MarkoMajstorovic the standard notation measures angles anticlockwise from the x axis. That is an arbitrary decision, but the definition of "cos" and "sin" then mean "use cos for the x axis and sin for the y axis". Note, the angle convention also works for graphing complex numbers. The formula $e^{i\theta} = \cos\theta + i\sin\theta$ is true by definition, so if you graph complex numbers with the real axis horizontal and the imaginary axis vertical, that fixes how $\theta$ is defined (i.e. measure it anticlockwise from the x axis). Apr 21 at 12:53

So to clarify, you're just trying to

1. plot the vector using it’s x-y components, or
2. or are you asking how to get the $$X$$ or $$Y$$ components of a vector?

If it’s # 1 it seems like you already did it since there’s no $$X$$ component and the $$Y$$ component is in the negative direction. If it’s # 2 then you're probably looking for $$x=R\cos\theta$$ and $$y = R\sin\theta$$ where $$R$$ is the force vectors magnitude and theta is the angle between the vector and the $$X$$ axis.

• Its the first one, but my question is more of a trigonometric one, why didnt we use sin(270) for the x axis of the magnitude $F_{ex}$ Apr 21 at 4:21
• @MarkoMajstorovic imagine a right triangle with its bottom side on the x-axis. That bottom side is the adjacent and the long side is the hypotenuse. The other end is the opposite, now you can see since it’s aligned, the adjacent is the x and the opposite is the y. The right triangle forms an angle theta with the x axis and by definition sin(theta) =opposite/hypotenuse or y/r Apr 21 at 4:31

Recall the triangle law of vectors.

To find the x and y components, construct a triangle with the vector $$F_e$$ as the hypotenuse.

This is where trigonometry comes in- since $$F_{ex}$$ is the base of the triangle and $$F_e$$ is the hypotenuse, you can now figure out the trigonometric relation between $$\theta$$, $$F_e$$ and $$F_{ex}$$.

• I mean this is not a homework question more of a how did they get that type of question, and how can i make Fe to be the hypotenuse ? Apr 21 at 4:36
• @MarkoMajstorovic in regular triangles the hypotenuse will be made in a regular form, but in this case it will overlap with y-axis(as you can see) and the x-axis will be reduced to zero.
– Sid
Apr 21 at 4:42
• To visualise: Imagine a regular right-angled triangle with $\theta = 45 \circ$. Now increase the angle between x-axis and hypotenuse, while keeping the hypotenuse constant. You can see that the length of the base would decrease, right? And at 90 deg., it would be just a line along the y-axis. Increasing it further, the triangle would go into the second quadrant.At 180 deg., it would be just a straight line along the negative x- axis. At 270, just a straight line along negative y-axis.
– Sid
Apr 21 at 4:47
• @MarkoMajstorovic mathsisfun.com/algebra/trig-interactive-unit-circle.html is a good visualisation
– Sid
Apr 21 at 4:56