How Do You Prove Trigonometric Identities for Vector Angles?

[Oblio]

First, thanks for all the help so far everyone!

vectors a and b exist in the x,y plane and make angles (alpha) and (beta) with x.

(Ill use A as alpha and B as beta)

prove: cos (A-B) = cos(A)cos(B)+sin(A)sin(B)

prove: sin (A-B) = sin(A)cos(B) - cos(A)sin(B)


I think there is some relationship (for the first one) to this:
r.s= rscos(theta), but I really don't know where to start with this one..

 

[learningphysics]

Hint: what is the angle between vectors a and b ?

Use dot product to get cos(A-B)... Use cross product to get sin(A-B).

 

[Oblio]

The angle is some angle (alpha) I guess. I suppose it doesn't matter where you apply each symbol.

 

[learningphysics]

The angle is some angle (alpha) I guess. I suppose it doesn't matter where you apply each symbol.

Yes, but what is the angle between the two vectors? If one vector makes an angle A with the x-axis. The other vector makes an angle B... what is the angle between the two vectors.

draw a picture.

 

[Oblio]

use the dot product on cos(A)cos(B)+sin(A)sin(B) you mean?

 

[learningphysics]

use the dot product on cos(A)cos(B)+sin(A)sin(B) you mean?

No, on the two vectors a and b. Use a.b = |a||b|costheta

What is the angle theta between the two vectors?

 

[Oblio]

the angle between them would be 180 - (the absolute value of alpha and beta)

 

[learningphysics]

the angle between them would be 180 - (the absolute value of alpha and beta)

Did they give a picture... if you measure the angles counterclockwise from the positive x axis... then the angle would be A-B (assuming A>B).

 

[Oblio]

no they didn't provide a picture...
when I drew it I drew (just randomly) alpha in the positive x section and beta in the negative x section. I drew an angle down to the x on both ( so the angle between crosses through the y-axis).

 

[learningphysics]

no they didn't provide a picture...
when I drew it I drew (just randomly) alpha in the positive x section and beta in the negative x section. I drew an angle down to the x on both ( so the angle between crosses through the y-axis).

Ah... I see... draw them both from the positive x direction make A the bigger angle... then A-B is the angle between them.

 

[Oblio]

ok, lol that seems like a vague question. so A-B is my angle giving:

lal . lbl = lal.lbl cos (A-B)

 

[learningphysics]

ok, lol that seems like a vague question. so A-B is my angle giving:

lal . lbl = lal.lbl cos (A-B)

it should be:

a . b = lal.lbl cos (A-B)

try to write a in the form (x,y)... use cosA sinA etc... along with |a|...

then try the same with b... but using cosB sinB and |b|...

 

[Oblio]

am i considering each vector independently or together?

i.e. vector a and its angle A to x, or vector a and the angle (A-B) ?

 

[learningphysics]

am i considering each vector independently or together?

i.e. vector a and its angle A to x, or vector a and the angle (A-B) ?

to get the x-component and y-component of vector a... so you'd use the angle A to x.

 

[Oblio]

so, like before I have extra info...
using the equation: (vectors) a.b = axbx + ayby
I substituted:

cosA(a)cosB(b)+ sinA(a)sinB(b)

I'm close to the answer...

 

[learningphysics]

so, like before I have extra info...
using the equation: (vectors) a.b = axbx + ayby
I substituted:

cosA(a)cosB(b)+ sinA(a)sinB(b)

I'm close to the answer...

yes, very you're close... to be precise you should use :
ax = |a|cosA, ay = |a|sinA, bx = |b|cosB, by = |b|sinB.

and get a.b using those...

then substitute your expression into the left side of:
a.b = |a||b|cos(A-B)

then a little simplification and you get the result.

 

[Oblio]

doesn't a.b = cosA(a)cosB(b)+ sinA(a)sinB(b) ?

 

[learningphysics]

doesn't a.b = cosA(a)cosB(b)+ sinA(a)sinB(b) ?

remember a and b are vectors... we need magnitudes on the right hand side.

The x-component of a vector is the magnitude of the vector times costheta from the positive x-axis.

The y-component of a vector is the magnitude of the vector times sintheta from the positive x-axis.

I'll give an example... take the vector (-3,4)... It has a magnitude of 5, and an angle with the positive x-axis of 143.1

so (-3,4) = (5cos143.1, 5sin143.1)

take another vector (1,1) = (1.41cos45, 1.41sin45)

(-3,4).(1,1) =
(5cos143.1,5sin143.1).(1.41cos45,1.41sin45) =

5cos143.1*1.41cos45 + 5sin143.1*1.41sin45

The reason I did this example, is to show that it is the magnitudes 5 and 1.41 that appear in the formula...

So a = (|a|cosA,|a|sinA)
b = (|b|cosB,|b|sinB)

so:

a.b = |a|cosA|b|cosB + |a|sinA|b|sinB

 

[Oblio]

comprendez.

Thanks a lot man

 

[learningphysics]

comprendez.

Thanks a lot man

no prob.

 

[learningphysics]

don't forget about the sin (A-B) = sin(A)cos(B) - cos(A)sin(B) part...