Hello math community,
i am studying for math at the moment and I am stuck a bit with the derivation of the law of cosines. First of all, sorry if i mention some things wrong, I study in german so it could be that we name some things different. :-)
My problem is that i don't know how to go on after a certain step. It would be cool if you could point me in the right direction .
Problem:
Derive the law of cosines from a triangle made of 2 vectors
a = { ax, ay az }
b = { bx, by, bz }
c = b - a
All lengths are unknown and arbitrary.
So, we are looking for c, alpha, beta & gamma
alpha is between c and b
beta is between c and a
gamma is between a and b
the longest side is c which can be separated into p and q which will give me two right angled triangles.
From this information i can easily get
cos alpha = p / b -> p = b * cos alpha
cos beta = q / a -> q = a * cos beta
c = p + q
c = b * cos alpha + a * cos beta
so far so good ^-^
In the book is written that I shall do the same for the sides a and b. And that is were i get stuck. All the time i am calculating the same stuff on and on again but i am not able to go one step further. It seems that i am missing some variable or s.th.
Lets say the opposite leg of alpha is h
so b = sqrt ( p * p + h * h )
p = cos alpha * b
h = ??? <- how
i ??
The goal is to get
c*c = a*a + b*b - 2ab*cos gamma
Angle between two vectors - the law of cosines - derivation problem
I'm not sure if this is correct, so please double-check:
b = sqrt(p*p + h*h)
b^2 = p^2 + h^2
b^2 - p^2 = h^2
h = sqrt(b^2 - p^2)
h = sqrt(b^2 - (cos(alpha) * b)^2)
h = b * sqrt(1 - (cos(alpha))^2)
h = b * sqrt((sin(alpha))^2)
h = b * sin(alpha)
b = sqrt(p*p + h*h)
b^2 = p^2 + h^2
b^2 - p^2 = h^2
h = sqrt(b^2 - p^2)
h = sqrt(b^2 - (cos(alpha) * b)^2)
h = b * sqrt(1 - (cos(alpha))^2)
h = b * sqrt((sin(alpha))^2)
h = b * sin(alpha)
Since you are working with vectors perhaps you should take a look at using the dot product, i.e. a.b = |a||b| cos(theta)
Hey , thanks for both answers..
I had as well the solution with h = b * sin alpha but the challenge is to do it without the sin :-)
Also no dot products ...:-(
But thanks a lot for your answers!!!!
I had as well the solution with h = b * sin alpha but the challenge is to do it without the sin :-)
Also no dot products ...:-(
But thanks a lot for your answers!!!!
Law of cosines states so I cannot see what's wrong with using
Anyways, your c = b * cos alpha + a * cos beta looks like Trigonometry proof from Wikipedia (source: http://en.wikipedia.org/wiki/Law_of_cosines#Using_trigonometry).
Anyways, your c = b * cos alpha + a * cos beta looks like Trigonometry proof from Wikipedia (source: http://en.wikipedia.org/wiki/Law_of_cosines#Using_trigonometry).
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