# Proving trig identities with dot and cross products

1. Aug 26, 2012

### cytochrome

1. The problem statement, all variables and given/known data

The two vectors a and b lie in the xy plane and make angles α and β with the x axis.

a)By evaluating ab in two ways (Namely ab = abcos(θ) and ab = a1b1+a2b2) prove the well-known trig identity
cos(α-β)=cos(α)cos(β)+sin(α)sin(β)

b)By similarly evaluating a X b prove that
sin(α-β) = sin(α)cos(β) - cos(α)sin(β)

c)Now let vector a make an angle -α with the x axis and find a similar expression for
cos(α+β)

2. Relevant equations

ab = abcos(θ)

ab = a1b1+a2b2

3. The attempt at a solution

I drew the vectors a and b with their appropriate angles to the x-axis... The angle between the vectors is (α-β) so I have ab = abcos(α-β) but I have no idea how to relate this to the trig identities!

2. Aug 26, 2012

### kushan

Hello ,
Since you are stuck at a.b=|a||b|cos(α-β) ,
Why dont you take use of the other equation you know a • b = a1b1+a2b2
a1 , a2 are components along i and j . Does that ring any bell?

3. Aug 26, 2012

### cytochrome

Yes that makes sense but I'm not sure how to relate that to the trig identity :/

4. Aug 26, 2012

### kushan

can you write a1 ,a2 , b1 ,b2 in terms of |a| or |b| and some trigo function ?

5. Aug 26, 2012

### cytochrome

a1=ax*cos(α)

a2=ay*sin(α)

b1=bx*cos(β)

b2=by*sin(β)

I'm currently playing around with these but any headers would be appreciated!

6. Aug 26, 2012

### kushan

you have a vector in terms of i , j just do the dot product and equate it two what you already have from equation 1 .
NOTE: you have posted this thread in multiple sections , moderators wont like it .

7. Aug 26, 2012

### cytochrome

Thanks.

Sorry I didn't know where to put this...