Proof of sin(θ-Φ)=sinθcosΦ-cosθsinΦ using vector algebra

1. Nov 7, 2015

Jamiemma1995

1. The problem statement, all variables and given/known data given two unit vectors a= cosθi + sinθi b=cosΦi+sinΦj prove that sin(θ-Φ)=sinθcosΦ-cosΦsinθ using vector algebra

2. Relevant equations sin(θ-Φ)=sinθcosΦ-cosΦsinθ

3. The attempt at a solution axb= (cosθsinΦ-cosΦsinθ)k and I'm guessing that the change in sign has something to do with the fact that k is perpendicular to the vectors I'm usingΦ-θ

2. Nov 7, 2015

haruspex

You could try a suitable rotation of axes, so that the cross product should still look the same.

3. Nov 7, 2015

ehild

The vectors a and b have zero "z" component. Remember how the components of the cross product are calculated.

4. Nov 8, 2015

Jamiemma1995

when I calculated the components I got axb = ( cosθi +sinθj )x(cosΦi + sinΦj) axb=cosθcosΦixi + cosθsinΦixj +sinθcosΦjxi +sinθsinΦjxj ixi=1x1xsino=0 jxjxsin0=0 ixj=1x1sin90=1 and jxi=-1 because AxB=-BxA and was then left with axb= cosθsinΦ(1) + sinθcosΦ(-1)=cosθsinΦ -sinθcosΦ but its supposed to be the other way around, I dont understand where I'm going wrong :(

5. Nov 8, 2015

ehild

Think of the other definition of the cross product, how to get its direction applying right-had rule, so sin(θ-φ)=bxa.
https://www.mathsisfun.com/algebra/vectors-cross-product.html

6. Nov 8, 2015

Jamiemma1995

oh I think I get it now so what your saying is although I could find axb and get the answer I got first , I could just as easily choose bxa as they're both the same but have opposite signs and so it still satisfies the proof.

7. Nov 8, 2015

ehild

The correct one is bxa.

8. Nov 8, 2015

J Hann

Isn't (i + j) X (i + j) = j X i + i X j)
What is the sign of j X i ?

9. Nov 8, 2015

ehild

minus. Why do you ask? It was correct in Post#4.

Last edited: Nov 8, 2015