Question

Suppose that vand ware unit vectors. If the angle between vand iis $\alpha$ and that between wand iis $\beta ,$ use the idea of the dot product $\mathbf{v}\mathbf{·}\mathbf{w}$to prove that

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

Short answer

It is proved that$\cos \left(\alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta$.

Step-by-step solution

Step 1. Given Information

there are two unit vectors vand w.The angle between vand iis $\alpha$ and that between wand iis $\beta .$ We have to use the idea of dot product and prove that
$\cos \left(\alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta .$

Step 2. Proving

If vand ware unit vectors.

So, $\left|\left|\mathit{v}\right|\right|=1,\left|\left|\mathit{w}\right|\right|=1.$

Now, the dot product of

$$\begin{gathered} \mathbf{v}\mathbf{·}\mathbf{w}=\left|\left|\mathbf{v}\right|\right|\left|\left|\mathbf{w}\right|\right|\mathrm{cos\theta } \\ \mathbf{v}\mathbf{·}\mathbf{w}=\mathrm{cos\theta }\left[\left|\left|\mathbf{v}\right|\right|=1,\left|\left|\mathbf{w}\right|\right|=1\right] \end{gathered}$$

When $\theta =\alpha -\beta or\beta -\alpha$we get,

$$\begin{gathered} \mathbf{v}\mathbf{·}\mathbf{w}=\mathrm{cos\theta } \\ \mathbf{v}\mathbf{·}\mathbf{w}=\cos \left(\mathrm{\alpha }-\mathrm{\beta }\right)........\left(\mathrm{a}\right) \end{gathered}$$

Now,

$$\begin{gathered} \mathbf{v}=\left|\left|\mathbf{v}\right|\right|\left(\mathrm{cos\alpha }\mathbf{i}+\mathrm{sin\alpha }\mathbf{j}\right) \\ \mathbf{v}=\left(\mathrm{cos\alpha }\mathbf{i}+\mathrm{sin\alpha }\mathbf{j}\right)\left[\left|\left|\mathbf{v}\right|\right|=1\right] \end{gathered}$$

And

$$\begin{gathered} \mathbf{w}=\left|\left|\mathbf{w}\right|\right|\left(\cos \beta \mathbf{i}+\sin \beta \mathbf{j}\right) \\ \mathbf{w}=\left(\cos \beta \mathbf{i}+\sin \beta \mathbf{j}\right)\left[\left|\left|\mathbf{w}\right|\right|=1\right] \end{gathered}$$

So, $$\begin{gathered} \cos \left(\alpha -\beta \right)=\mathbf{v}\mathbf{·}\mathbf{w} \\ \cos \left(\mathrm{\alpha }-\mathrm{\beta }\right)=\left(\mathrm{cos\alpha }\mathbf{i}+\mathrm{sin\alpha }\mathbf{j}\right)·\left(\mathrm{cos\beta }\mathbf{i}+\mathrm{sin\beta }\mathbf{j}\right) \\ \cos \left(\mathrm{\alpha }-\mathrm{\beta }\right)=\mathrm{cos\alpha }\mathrm{cos\beta }+\mathrm{sin\alpha }\mathrm{sin\beta } \end{gathered}$$

Hence proved.