Question

Why does trigonometric function works in solving vector force if trig functions are define in length not force?
why does trig function applies to different magnitudes if it is define in length?
Fx = F cos(theta)?
but cos(theta)=b/c
and b and c is by definition in magnitude of length not force
When drawing force diagrams, the LENGTH of the force represents its magnitude.
but it only represents the force but not the force itself
length is not force

what is the definition of trigonometric functions?
is it defined as a function of an angle and ratio of the LENGTH of two sides?
or is it defined as a function of an angle and ratio of the magnitude(force,velocity,length.....abstract magnitude) of the two sides?
and suppose it is the definition as ratio of LENGTH how come that it also applies for other magnitudes?
furthermore what is Pythagorean theorem?
the mathematical proof of the theorem implies that it can solve the hypotenuse of the right triangle given the LENGTH of the two sides.The proof assumes that the magnitude is LENGTH and not of other magnitude.
Am I right or just confuse?
i don't know

Answer 1

Yes. what Waynex said is correct. It is better not to consider vector as having a length, they do not. They have a magnitude (a scalar value) and a direction. The trigonometric functions helps you to find the scalar values, or the magnitudes. They do not help you to find "lengths."

Answer 2

what is the definition of trigonometric functions?
is it defined as a function of an angle and ratio of the LENGTH of two sides?
or is it defined as a function of an angle and ratio of the magnitude(force,velocity,length.....abstract magnitude) of the two sides?
and suppose it is the definition as ratio of LENGTH how come that it also applies for other magnitudes?
furthermore what is Pythagorean theorem?
the mathematical proof of the theorem implies that it can solve the hypotenuse of the right triangle given the LENGTH of the two sides.The proof assumes that the magnitude is LENGTH and not of other magnitude.
Am I right or just confuse?
i don't know

Answer 3

here is what i think(i'll explain the "velocity" version proove but you'll get the idea):

if you say two velocity(vx vy)is vertical it means that after t seconds ,their displacements is vertical that is: dx=vxt is vertical to dy=vyt

therefore: \[ \left( vt \right) ^{2}=\left( v _{x}t \right)^{2}+\left( v _{y} t\right)^{2}\]
since vxt is vertical to vyt and vt vxt vyt are all lengths

and if you eliminate t you'll get:
\[v ^{2}=v _{x}^{2}+v _{y}^{2}\]

Answer 4

I think I have found the solution but anyway thanks for all your answers.
the idea is to find an alternative proof of Pythagorean theorem in such a way that it incorporates other magnitude(force,e-field,velocity)that is,instead of the statement "The square of the LENGTH of the hypotenuse equals the sum of the squares of the LENGTHS of the other two sides." we need to find a proof that will lead to the conclusion of "The square of the MAGNITUDE of the hypotenuse equals the sum of the squares of the MAGNITUDES of the other two sides."
In order to do that we use vector to derive the Pythagorean theorem by using dot product,properties of vector,unit vector.
suppose we have a vector V as illustrated below and we want to find its magnitude.
The components are A(x-coordinate) and B(y-coordinate).

Now rotate the figure above in a way that V become horizontal.

The magnitude of the vectors did not change but the direction is.

Let D be the new transform V,K be the new transform A and L be the new transform B.


Using coordinate system,D could be translated as:
since |D| = |C| then
D=|V|*i+0*j or D=<|V|,0>
where i(x-coordinate) and j(y-coordinate) are unit vectors.


By definition of Dot product:

D.D = |V|*|V| + 0*0

since |K|=|A| and |L| = |B|
then
D=<|A|,|B|>

D.D =|A|*|A|+|B|*|B|

therefore
|V|*|V| + 0*0=|A|*|A|+|B|*|B|
|V|^2=|A|^2+|B|^2
but V.V=|A|^2+|B|^2 because V=<<|A|,|B|>
we further conclude that :
|V|^2=V.V or |V|=square_root(V.V)

The next step is to use the generalized Pythagorean theorem to define unit circle for trigonometric functions

sin0=Y/square_root(X^2 + Y^2)
cos0=X/square_root(X^2 + Y^2)
but since |V|^2= |A|^2 + |B|^2 is equal to C^2= X^2 + Y^2
then
sin0=|B|/square_root(|A|^2 + |B|^2)
cos0=|A|/square_root(|A|^2 + |B|^2)
tan0......and so on

this proves that the sin0 in terms of magnitude is equivalent from sin0 in terms of length and leading to the conclusion that the trig functions and Pythagorean theorem equally applies to other magnitudes(force,e-field,velocity etc...)

Answer 5

I think it is a bit strange,why does D=<|A|,|B|> and <|V|,0>?

Answer 6

the magnitude of K and L is equivalent to |A| and |B|

another interpretation of the figure above is