Chapter C, Problem 43E

To determine

To prove: The subtraction formula $\cos \left(\alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta$ by the use of given figure.

Explanation of Solution

From the given figure, it is observed that $A\left(\cos \alpha ,\sin \alpha \right)\text{and}B\left(\cos \beta ,\sin \beta \right)$ are two distinct point and c is the distance from A to B.

Obtain the value of $c^2$ by law of cosine.

$\begin{array}{c}{c}^2=1^2+1^2-2\left(1\right)\left(1\right)\cos \left(\alpha -\beta \right)\\ =2\left(1-\cos \left(\alpha -\beta \right)\right)\end{array}$

That is, $c^2=2\left(1-\cos \left(\alpha -\beta \right)\right)$ (1)

The distance between any two point $\left(x_1,x_2\right)\text{and}\left(y_1,y_2\right)$ is $\sqrt{{\left(x_1-y_1\right)}^2+{\left(x_2-y_2\right)}^2}$.

Thus, the distance c from $\left(\cos \alpha ,\sin \alpha \right)$ to $\left(\cos \beta ,\sin \beta \right)$ is $c=\sqrt{{\left(\cos \alpha -\cos \beta \right)}^2+{\left(\sin \alpha -\sin \beta \right)}^2}$.

Further simplified as,

$\begin{array}{c}{c}^2={\left(\cos \alpha -\cos \beta \right)}^2+{\left(\sin \alpha -\sin \beta \right)}^2\\ ={\cos }^2\alpha +{\cos }^2\beta -2\cos \alpha \cos \beta +{\sin }^2\alpha +{\sin }^2\beta -2\sin \alpha \sin \beta \\ =\left({\cos }^2\alpha +{\sin }^2\alpha \right)+\left({\cos }^2\beta +{\sin }^2\beta \right)-2\cos \alpha \cos \beta -2\sin \alpha \sin \beta \\ =2-2\cos \alpha \cos \beta -2\sin \alpha \sin \beta \end{array}$

That is, $c^2=2-2\cos \alpha \cos \beta -2\sin \alpha \sin \beta$ (2)

From equation (1) and (2) it is observed that,

$2\left(1-\cos \left(\alpha -\beta \right)\right)=2-2\cos \alpha \cos \beta -2\sin \alpha \sin \beta$

Further simplified as,

$\begin{array}{l}2\left(1-\cos \left(\alpha -\beta \right)\right)=2-2\cos \alpha \cos \beta -2\sin \alpha \sin \beta \\ 2\left(1-\cos \left(\alpha -\beta \right)\right)=2\left(1-\cos \alpha \cos \beta -\sin \alpha \sin \beta \right)\\ 1-\cos \left(\alpha -\beta \right)=1-\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ \cos \left(\alpha -\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta \end{array}$

That is, $\cos \left(\alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta$.

Hence the proof.