# How do we get a cos component and a sin component when we resolve a vector?

So, I'm trying to understand this.

In the following we've an image, with a vector $v_0$ and we can see the components of the vectors have trigonometric ratios such as $\sin\theta$ and $\cos\theta$ for vertical and horizontal components respectively.

But how can we proof, that the vertical component be $v_0\sin\theta$ and horizontal component be $v_0\cos\theta$?

I know that when we add two vectors, the magnitude of the resultant vector is $$=\sqrt {a^2+b^2+2ab\cos \theta}$$ through the triangle law, but how can we prove this?

• How much trigonometry do you know? Given a right angled triangle, have you seen something which looks like $\sin(\theta) = \frac{O}{H}$ before?
– Matt
Aug 26, 2018 at 14:27
• @Matt if you're implying the perpendicular by O, then yes. Aug 26, 2018 at 14:28
• It is basically just the definition: you are given the hypotenuse is length $v_0$ and you are given the acute angle $\theta$, so you extract the lengths of the legs in terms of $\cos \theta$ and $\sin \theta$.
– Ian
Aug 26, 2018 at 14:30
• Well, here you have an angle $\theta$, and $H = v_0$. You know that $\sin(\theta) = \frac{O}{H}$, and also that $\cos(\theta)=\frac{A}{H}$. You can rearrange both of these equations to find expressions for $A$ and $O$. Can you see how this fits in with your diagram?
– Matt
Aug 26, 2018 at 14:31
• oh yeah, i totally forgot. It's a vector, a vector implies both magnitude and direction not points in a space. hence we can arrange the vertical component in an order that it becomes the perpendicular / opposite of the triangle and then using the the ratios, we can say that the vertical component is v0sinθ. Aug 26, 2018 at 14:32

Thus the components are the norm multiplied by sine and cosine of the angle between your vector and the positive direction of the $x$-axis.