Chapter 9.5, Problem 37AYU

To determine

To prove: $\cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta$

Answer to Problem 37AYU

Solution:

  $\cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta$

Explanation of Solution

Given:

  $v$ and $w$ are unit vectors. If the angle between $v$ and $i$ is $\alpha$ and the angle between $w$ and $i$ is $\beta$

Calculation:

We know, if $v$ is a unit vector and the angle between $v$ and $i$ is $\alpha ,$ then $v=\cos \alpha i+\sin \alpha j$

If the angle between $v$ and $i$ is $\alpha$ and the angle between $w$ and $i$ is $\beta$ .

Therefore $v=\cos \alpha i+\sin \alpha j$ and $w=\cos \beta i+\sin \beta j$

since $v$ and $w$ are unit vectors $\Vert v\Vert =1$ and $\Vert w\Vert =1$ .

  

  $\cos \theta =\frac{v\cdot w}{\Vert v\Vert \Vert w\Vert }$

  $\cos (\beta -\alpha )=\frac{(\cos \alpha t+\sin \alpha j)(\cos \beta i+\sin \beta j)}{1}$

  $\cos (\beta -\alpha )=\cos \alpha \cos \beta +\sin \alpha \sin \beta$

  $\cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta ,$ since cosine is an even function.