Chapter 11.3, Problem 69E

To determine

To proof: The trigonometric identity $\sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$ with help of cross product of vectors $u=\langle \cos \alpha ,\sin \alpha ,0\rangle \text{ and }v=\langle \cos \beta ,\sin \beta ,0\rangle$ where $\alpha >\beta$ .

Explanation of Solution

Given information:

The vectors $u=\langle \cos \alpha ,\sin \alpha ,0\rangle \text{ and }v=\langle \cos \beta ,\sin \beta ,0\rangle$ .

Formula used:

If there are two three-dimensional vectors in a space given in the form of their components, say, $a=\langle a_1,a_2,a_3\rangle$ and $b=\langle b_1,b_2,b_3\rangle$ , then the cross product of two vectors is defined as ,

  $\begin{array}{c}a\times b=\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\\ \text{= }\left(a_2{b}_3-a_3{b}_2\right)i-\left(a_1{b}_3-a_3{b}_1\right)j+\left(a_1{b}_2-a_2{b}_1\right)k\end{array}$

Let $a$ and $b$ be any two arbitrary vectors, then the cross product of the vectors is denoted by $a\times b$ .

Mathematically the magnitude of cross product is expressed as, $\Vert a\times b\Vert =\Vert a\Vert \cdot \Vert b\Vert \cdot \sin \theta$ , where $\Vert a\Vert$ and $\Vert b\Vert$ denote the magnitude of vectors $a$ and $b$ respectively and $\theta$ is the angle between them.

Proof:

Consider the set of vectors $u=\langle \cos \alpha ,\sin \alpha ,0\rangle \text{ and }v=\langle \cos \beta ,\sin \beta ,0\rangle$ where $\alpha >\beta$ .

Recall that if there are two three-dimensional vectors in a space given in the form of their components, say, $a=\langle a_1,a_2,a_3\rangle$ and $b=\langle b_1,b_2,b_3\rangle$ , then the cross product of two vectors is defined as ,

  $\begin{array}{c}a\times b=\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\\ \text{= }\left(a_2{b}_3-a_3{b}_2\right)i-\left(a_1{b}_3-a_3{b}_1\right)j+\left(a_1{b}_2-a_2{b}_1\right)k\end{array}$

Apply it,

  $\begin{array}{c}u\times v=\begin{array}{ccc}i& j& k\\ \cos \alpha & \sin \alpha & 0\\ \cos \beta & \sin \beta & 0\end{array}\\ \text{= }\left(0-0\right)i-\left(0-0\right)j+\left(\cos \alpha \sin \beta -\sin \alpha \cos \beta \right)k\\ =-\left(\sin \alpha \cos \beta -\cos \alpha \sin \beta \right)k\end{array}$

The magnitude of $u\times v$ is,

  $\begin{array}{c}\Vert u\times v\Vert =\sqrt{{\left[-\left(\sin \alpha \cos \beta -\cos \alpha \sin \beta \right)\right]}^2}\\ =\sin \alpha \cos \beta -\cos \alpha \sin \beta \end{array}$

Let $a$ and $b$ be any two arbitrary vectors, then the cross product of the vectors is denoted by $a\times b$ .

Mathematically the magnitude of cross product is expressed as, $\Vert a\times b\Vert =\Vert a\Vert \cdot \Vert b\Vert \cdot \sin \theta$ , where $\Vert a\Vert$ and $\Vert b\Vert$ denote the magnitude of vectors $a$ and $b$ respectively and $\theta$ is the angle between them.

  

The angle between the vectors $u=\langle \cos \alpha ,\sin \alpha ,0\rangle \text{ and }v=\langle \cos \beta ,\sin \beta ,0\rangle$ is $\theta =\alpha -\beta$ .

Now,

  $\begin{array}{c}\Vert u\times v\Vert =\Vert u\Vert \cdot \Vert v\Vert \cdot \sin \theta \\ =\sqrt{{\cos }^2\alpha +{\sin }^2\alpha +0}\sqrt{{\cos }^2\beta +{\sin }^2\beta +0}\left[\sin \left(\alpha -\beta \right)\right]\\ =\sqrt{1}\sqrt{1}\sin \left(\alpha -\beta \right)\\ =\sin \left(\alpha -\beta \right)\end{array}$

Equate the value of $\Vert u\times v\Vert$ obtained from both the methods, $\sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$ .

Hence, the identity $\sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$ is provedwith help of cross product of vectors $u=\langle \cos \alpha ,\sin \alpha ,0\rangle \text{ and }v=\langle \cos \beta ,\sin \beta ,0\rangle$ where $\alpha >\beta$ .