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I have always wondered why you have to use sine and cosine instead of a proportional relationship, such as $(90-\text{angle})/90$. I cannot seem to find an answer anywhere online. Using my linear relationship, when the angle is $0$, then $90/90$ is $1$ and the component is at its maximum value, and when the angle is $90$, the component is $0$, both of which are true for cosine. But how come in between those values, there is a relationship of cosine?

For example, if there is a vector of $5$ at an angle $35$ degrees to the horizontal, why does the horizontal component have to be $5\cos 35^{\circ}$, why can't it be $5 \cdot (90-35)/90$?

Also, this leads me on to another question. Using the same example, if the vector is $5$ at $35$ degrees, then how can the sum of the horizontal and vertical components be greater than the original $5$? $5\cos 35^{\circ} + 5\sin 35^{\circ} = 6.96$ which is greater than $5$? Magic? Energy crisis solved?

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The vector is the diagonal of a unique rectangle. The horizontal and vertical components of the vector are by definition the horizontal and vertical sides of the rectangle; that's just the statement that moving from the origin to the endpoint along the ray gives the same ending location as moving only horizontally and then only vertically along those sides.

Your ratio is measuring the fraction of the arc along the circle above or below the endpoint. The factor of the radius ($5$) is meaningless here and only muddles the expression more.

The definition of the components should make it clear why they add up to more than the length of the original vector. It's essentially that the shortest path between two points is a straight line (so the diagonal is shorter than the sum of the two edges). Not so surprising.

Note that your measure is also inherently flawed for the following reason. If you start with a vector that is horizontal and rotate it up a little bit, the change is almost entirely in the vertical direction and very little in the horizontal direction. Likewise if you start with a vector that is vertical and rotate it down a little bit, the change is almost entirely in the horizontal direction and very little in the vertical direction. But your measurement would give a change that is the same for both. The sine and cosine, on the other hand, take this into account.

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  • $\begingroup$ This somewhat answers the question, its the closest to what im looking for out of all the current answers, especially your last paragraph $\endgroup$ – Ethan SK May 7 '17 at 15:20
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Because of the Pythagorean theorem.

In more detail: Consider a vector 2 units long, pointing at a 45 degree angle from the horizontal. According to your proposal, the horizontal and vertical components would each be 1. But that would mean we have a right triangle with sides equal to 1, 1, and 2. Do you see why that would violate the Pythagorean theorem?

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  • $\begingroup$ Yes, I know it is wrong in practice, but that is not my question. My question is why is it wrong. Why cant the triangle have sides 1, 1, and 2? I know they cannot in real life, but what is the maths behind having to use sin and cos $\endgroup$ – Ethan SK May 7 '17 at 15:17
  • $\begingroup$ @EthanSK My answer does not say anything about sin and cos. Is your question "Why is the Pythagorean theorem true?" $\endgroup$ – mweiss May 7 '17 at 16:03
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Ok, I'll start with the second question.

Fact: For any triangle in the world with sides $a,b,c$: $$ |a-b| < c < a+b $$ So there is no surprise that the sum of the components is bigger than the length of the vector. If you don't understand which triangle I'm talking about, see this image:

enter image description here

Why do you think it is related to energy? This is a mathematical concept for vectors and energy is a scalar.

About the first question, it just comes from the geometrical definitions of sine and cosine, are you aware of them?

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  • $\begingroup$ I am aware of them, I want to know why. Essentially proof. $\endgroup$ – Ethan SK May 7 '17 at 15:15
  • $\begingroup$ Look at the right triangle above with sides A,Ax,Ay. What will be, by definition, $\sin \theta$? what will be $\cos \theta$? $\endgroup$ – Ofek Gillon May 7 '17 at 15:24
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We use the sine and cosine to decompose a vector into its horizontal and vertical components because the definition of sine and cosine (geometric wise) allows us to do so. enter image description here

Consider the diagram below. By definition of cosine we have:

$cos \theta= \frac{|v_x|}{|v|}$ and so $|v_x|=|v|cos \theta$.

Similarly, $|v_y|=|v|sin \theta$.

Note that the expression merely gives the magnitude of the horizontal and vertical components of the vector itself.

So to answer your last question, we don't add vectors like 5cos35 + 5sin35 = 5. Instead by the pythagoras theorem, $|v|^2=|v_x|^2 + |v_y|^2$ which you can verify to be true by virtue of the identity $sin^2\theta+cos^2\theta=1$.

It is not by magic and sadly, the energy crisis is still ongoing and cannot be solved that easily.

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The reason we use sine and cosine is because of the way they are defined for triangles. Remember that for an angle $\theta$ in a triangle, \begin{equation*} \sin\theta = \frac{\text{length of opposite side}}{\text{length of hypotenuse}},\quad\cos\theta= \frac{\text{length of adjacent side}}{\text{length of hypotenuse}}. \end{equation*}

This fits naturally with vectors and finding their horizontal and vertical components. For you case, the length of the hypotenuse is $5$, and so we can say

$$ \sin 35^{\circ} = \frac{\text{length of vertical component}}{5} \implies \text{length of vertical component} = 5\sin35^{\circ} $$

and similarly $$ \cos 35^{\circ} = \frac{\text{length of horizontal component}}{5} \implies \text{length of horizontal component} = 5\cos 35^{\circ} $$

There's nothing magical happening with the sum of the horizontal and vertical components because if you add together the lengths of two sides of a triangle you always get something greater than the length of the remaining side.

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