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?

projectile motion, resolution of vectors

  • $\begingroup$ How much trigonometry do you know? Given a right angled triangle, have you seen something which looks like $\sin(\theta) = \frac{O}{H}$ before? $\endgroup$
    – Matt
    Aug 26, 2018 at 14:27
  • $\begingroup$ @Matt if you're implying the perpendicular by O, then yes. $\endgroup$
    – Daksh M.
    Aug 26, 2018 at 14:28
  • $\begingroup$ 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 $\sin \theta$. $\endgroup$
    – Ian
    Aug 26, 2018 at 14:30
  • $\begingroup$ 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? $\endgroup$
    – Matt
    Aug 26, 2018 at 14:31
  • $\begingroup$ 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. $\endgroup$
    – Daksh M.
    Aug 26, 2018 at 14:32

1 Answer 1


These results are coming from the definition of sine and cosine functions.

Your hypotenuse is the norm of your vector and the components are the adjacent and opposite sides of a right triangle.

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.

  • $\begingroup$ thanks, i'll accept this as an answer. $\endgroup$
    – Daksh M.
    Aug 26, 2018 at 14:39
  • $\begingroup$ @DakshMiglani great, thank you very much. $\endgroup$ Aug 26, 2018 at 14:40

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